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KINETIC THEORY OF MATTER
It is the theory that tries to explain how matter behaves. It is based on the following assumptions
1. Matter is made up of very tiny particle invisible with our naked eyes and to microscope.
2. The particles of matter are in constant random motion. The motion increase with temperature.
Small particle move faster than the large one
3. There are forces of attraction between the particles
State of matter
Matter exist in three states these are gas, liquid and solid
Solids
Properties of solids
1. They have fixed shape, volume , density
2. They are closely packed
3. They have strong IMF ( Intermolecular forces)
4. The particles vibrate on a fixed position
5. They have regular shape
Liquid
1. They fairly close to each other
2. They have quite strong IMF
3. They move in a straight line (rectilinear motion)
4. They take the shape of the container
Gases
1. Particles are far apart
2. They have week IMF
3. The move in straight kind at very high speed randomly
4. They have no any structure
Evidence of existence of particles
1. Diffusion
It is the movement of particles from high concentration to low concentration. It takes place in
haphazard and at a random way.
For example if the liquid of bromine is dropped into a jar containing air the jar is covered , the
liquid of bromine vaporise and after some times the bromine vapour can spread evenly through
he gas jar.
2. Diffusion of ammonia and sodium hydrogen chloride
Experiment 1
Evidence of existence of particles
Se the apparatus as show below
Procedure
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1. Hold a stopper of ammonia bottle near the open acid bottle. Observe what happens
2. Make sure that the glass tube is dry inside and out
3. Soak a pinch of cotton wool in ammonia solution and put it into the other end of the tube. Close the
end of the tube with other cork
4. At once soak the other in acid (hydrochloric acid) put it on the other end as showdown on the
diagram above and close the side with a cork
5. Place the tube direct sunlight on the bench ( or clamp it horizontally) and wait for moment.
Expected Results
A white solid will be formed. This will formed because the Ammonia reacts with an acid to form
ammonium chloride which is white in colour.
Discussion
Since the particles move and meet at a certain point. This provide example that particle move from
one point to another.
GAS PRESSURE
Pressure is defined as force per unit area
P= F/A
Gas pressure is measured in following units
1. Atmospheres ( atm)
2. Millimetre of mercury ( mm Hg)
3. Newton’s per square Metre called Pascal
1 atm = 1.0132 x 10
5
Nm
-2
= 1.0132X10
5
Pascal = 760mmHg
How temperature affect molecular motion
Temperature increase the kinetic energy since the heat energy is converted to kinetic energy in a
particle. Since the kinetic energy has increase this makes particles to move fasts than before. When
temperature increase molecule to move faster than before since heat energy is also increased which
increases the kinetic energy.
How pressure is caused
When air particle are closed in a contained. The particle move rapidly and randomly hitting the wall of
the container
According to kinetic theory of matter it states that matter has a kinetic energy. It also said that particle
in gas move randomly and in rapid speed. This causes the gas particle hit the wall of the container
frequently. This cause the pressure inside the container.
FACTORS THAT AFFECT GAS PRESSURE
a. Temperature
When the temperature is increase the kinetic energy of the particles also increases. This means
that the particle will be hitting the walls of the container more often than before. The
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temperature increases the kinetic energy in the gas particles which in turn increase pressure in
the container.
b. Volume of the container
The pressure of the gas increase and the volume of the container decreases. This mean that
particles will hit the walls of the container more often than before because the gases particle
will travel short distance to hit the walls of the container due to small space. In large container
particles travel long distance to hit the wall of the container whichreduces the number of
collision per seconds.
c. Number of particle in the container( concentration )
Increase the number of particles number of times the molecules will hit the walls per second.
TEMPERATURE
It is the measure of how hot or cold a substance is
Temperature scales
There two temperature scales that are commonly used in science and these are
1. The Celsius scale
It uses Degree Celsius on the scale.
2. Kelvin scale
It uses Kelvin and denoted by K
Temperature conversion
0°C = 273K
Change the follow in degree to Kelvin’s
15°C (b) 78 C°
a). K = 273 K + 15 = 298K
b). K = 273K + 78 = 351
Absolute temperature
It correspond to Kelvin scale of measuring temperature. In absolute temperature only positive value
are allowed. When temperature is zero it is considered that the quantity being measured( tempered) is
consider completely absent.
Absolute zero temperature
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It is the temperature which is considered no a single particle vibrates and all substance has no internal
kinetic energy. At this temperature, all substance has no internal energy and it correspond to 0K on
Kelvin scale and -273°C of Celsius scale.
GAS LAWS
Boyle's Laws
It is the law that tries to link the relationship between volume and pressure when temperature is kept
constant
It state that pressure of fixed mass of the gas is inversely proportional to its volume if the temperature
is kept constant.
Graphically
Volume α 1/pressure
V α 1/P
V = k/P where k is a constant of proportionality
therefore k = PV
P
1
V
1
= P
2
V
2
Example
If a certain mass of gas has the volume of 7.50 x10
-5
m
3
at a pressure of 15.0kpa. What will be the
volume of the gas if the pressure is altered to 25.0kpa without changing the temperature.
Given: P
1
= 15.kp
V
1
= 7.5 x10
-5
m
3
P
2
= 25.0Kpa
V
2
=
P
1
V
1
= P
2
V
2
15.0 x 7.5 x10
-
5
= 25.0 xV
2
V
2
= 15 x7.5 x 10
-5
25.0
= 4.50 x10
-5
Therefore the volume of the gas at 25kp is 4.5 x 10
-5
m
3
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Charles laws
It is the law that tries to link the relationship between volumeand temperature if temperature is kept
constant
It state that at fixed mass of gas,Volume is directly proportional it temperature if pressure is kept
constant,
Graphically
Mathematically
V α temperature
V α T
That is V = kt where k is constant
therefore V/T = k
In general
V
1
= V
2
T
1
T
2
Example:
Mr Kanyambita inflate the tire to 25kpa which to a volume of 500cm
3
at temperature of 300K. Later
in the day the temperature increase to 320K . find the volume of the tire at a new temperature?
Given:
V
1
= 500cm
3
T
1
= 300K T
2
= 320K V2= ?
V
1
= V
2
T
1
T
2
500 = V
2
300 320
V
2
= 500 x 320
300
V
2
= 533.33cm
3
Pressure law
This law link up the relationship between pressure, temperature if volume is kept constant. It state that
for a fixed mass of gas, pressure is directly proportional to its absolute temperature when the volume is
kept constant.
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Graphically
Pα T
P = kT where k is a proportionality constant
P/T = k
Therefore P
1
= P
2
T
1
T
2
Example: The pressure of the air in a bicycle tire is 240kpa at 300K. What will be the pressure in the
bicycle tire if the temperature is raised to 350K
Given
P
1
= 240k T
1
= 300
P
2
= ? T
2
= 350K
P
1
= P
2
T
2
T
1
240 = P2
300k 350
P
2
= 240 x 350
300
= 280kpa
Ideal gas law / General gas law
This combines three laws, Charles law, Boyle law and the pressure law. This is applicable only when
the gas is no liquefier because the intermolecular force influencesthe influence the behaviour of the
gases at that point.
The equation is used to determine changes when one or two things have changed.
P
1
V
1
= P
2
V
2
= Constant where
T
1
T
2
Example: The volume of the gas at 27°C with 320mmHg is 20 L. . Find the temperature when the
volume will be 30Land the pressure is 480mmHg
Given:
Original pressure = 320mmhg
Original volume = 20L
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Original temperature = 27°C
Final temperature = ?
Final pressure = 480mmHg
Final volume = 30L
320 x 20 = 480 x 30
27 T
T = 480x 30 x 27
320 x 20
= 60°C
Example 2: A balloon, volume 0.5m
3
, contain hydrogen at a pressure of 2x10
5
Pa is released from the
ground when the temperature is 17°C . What will be the volume when it reaches a height where the
pressure inside balloon is 1,5x 10
5
Pa and the temperature is 6°C
Given:
Original pressure = 2x10
5
Original volume =0,5m
3
Original temperature = 17°C
Final temperature = 6C
Final pressure = 1.5x10
5
Final volume = ?
2 x10
5
x 0.5 = 1.5 x 10
5
x V
2
17 6
2x10
5
x 0.5 x 6 = V
2
17 x 1.5 x10
5
0 .23m
3
= V
2
PRESSURE IN LIQUIDS
Properties of liquid pressure
Liquid pressure has the following properties
1. Pressure increase with depth because the farther you go the greater the weight of the liquid above
that exert force on the surface particles.
2. At certain depth of the liquid the pressure is the same at all point at that level and in all direction
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3. Liquid find its own level.
4. Pressure does not depend on the shape of the container
5. Any external pressure is transmitted to all parts of the liquid.
6. Pressure act at right angle to the container
Pressure formula in fluids
Pressure = F/ d
P = F /A
From the figure above
V = A x h where A is the area of cross section of the base
But density (d) = mass(M) / volume (v)
Mass of liquid = d x v = d xAxh
Recall that weight of substance = mg
Therefore mass of a substance is d. A. h .g
But the weight of the liquid is distributed over all parts of the liquid
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P = F = W = d. A .h .g
A A A
Therefore pressure (P) = d . h. g since A ( area) cancel each other . Where d is the density of liquid in
kg/m3 , h is the height of the liquid in the container, and g is the gravitational acceleration which is
equal to 10m/s/s
Example
What is the pressure containing on the base of the container 76 cm long containing Mercury with a
density of 13600kg/m
3.
Take g as 10m/s/s
Solution
Given:
Density = 13600kg/m
3
height of the container = 76cm
g = 10m/s/s
Pressure = d. h. g
P = 13600 x 0.76 x 10
= 103360pa
Uses Of Liquid Pressure
1.. Hydraulic Jack
It has a platform on top of piston B and is used in garages to lift cars. See the figure below.
B is raised a long way when A moves up and down repeatedly.
2. Hydraulic Car Brakes
When the brake pedal is pushed down , a small piston in a master cylinder forces brake fluid through
narrow pipes leading to the four wheels. Here the fluid pushes on pistons in slave cylinders, and the
pistons move brake pads or shoes against metal drums or discs attached to the wheels and stop the
car
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3. The knowledge of liquid pressure can be applied when designing a dam to calculate the pressure at
various depths below the water surface and making strong walls. Before building a dam, engineers
calculate the total force of water that will press against its face. To do this , they compute the
pressure half-way down from the surface, called average pressure .This average pressure
multiplied by the area of the whole face of the dam gives the total force exerted against it by the
water.
4. Water tanks are also placed at high altitude so that water goes to hoses or offices with great
pressure.
5. Pumps. When the tight fitting piston is drawn back along the barrel the pressure in the barrel is
reduced and the atmospheric pressure drive the liquid in and when the piston is hushed down it
drive the liquid out of the small nozzle at high pressure.
Siphons , the container containing liquid is placed is place top where the container where liquid will be
transferred is placed place below. The air in the tube is removed so to create a low pressure and is
replaced with the liquid while one end. After filled both ends is closed using a finger and dipped into
the tank and another end to the container. The pressure below the tank pushes the liquid out the tube to
the container below .
MANOMETER
It is instrument that can be used to measure differences in gas or liquid pressure. It is filled with
mercury. One end is Connected to the gas supply while the other is open to atmospheric pressure.
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It uses the principle that say that external pressure is transmitted to all parts of the liquid and that the
pressure in au table is the same at any point .
Each surface of the liquid is acted on equally by the atmospheric pressure and the levels are the same in
A.
I f one side is connected to , for example , the gas supply , the gas exerts a pressure on surface A and
level B rises up which can be measured by the ruler. The pressure difference pushes the mercury or the
liquid round the u tube and there is a height difference between mercury levels as a result as shown in
B.
Pressure of gas = atmospheric pressure + pressure due to liquid column BC.
The pressure of the liquid column BC equals the amount by which the gas pressure exceeds
atmospheric pressure. If for example, the height difference were 30 mm Hg and atmospheric pressure
were 760mm Hg, the pressure of the gas would be 790mmHg.
Note :
If the difference between the mercury levels is too small to measure accurately, water or oil can be
used in the manometer instead of mercury.
To find the pressure exerted on liquid use the formula P = d.h. g ( in Pa ) where h is the vertical
height of BC ( in m ) and d is the density of the liquid ( in kg /m
3
).
Measuring Lung Pressure Using Manometer
Lung pressure can be measured by connecting one end of the manometer to the mouth and try to
breathe out very hard. Air from the lungs can exert pressure on the liquid in the manometer. This can
cause difference in levels.
To find the pressure exerted on the liquid or Lung pressure = atmospheric pressure + pressure due to
the liquid column or use the following formula P = dhg and density should be in kg/m
3
Pascal s Law:
It states that pressure applied to a confined fluid acts equally in all directions.
Note: Hydraulic devices are the applications of Pascal s law.
The word hydraulic refers to any device that operates by applying pressure to a liquid.
Examples of hydraulic machines are:
1. Hydraulic presses for baling cotton, extracting oils from seeds, shaping automobile bodies,
punching holes in steel plates and moulding articles from plastics.
2. Hydraulic brakes on automobiles and airplanes
3. automobile lifts used in service stations.
4. Barber chairs.
5. Pneumatic devices are operated by a compressed gas.
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6. Can be applied in hydraulic machines such as hydraulic car jack and hydraulic car brakes.
Hydraulic machines work by using liquids under pressure rather than levers or wheels.
They make use of two properties:
a. Liquids are virtually incompressible;
b. If pressure is applied to a trapped liquid, the pressure is transmitted to all parts of the liquid.
Suppose a downward force of 1N acts on piston A of area 1/100m
2
. The pressure transmitted through
the liquid is :
Pressure = Force/Area = 1N / 1/ 100m
2
= 100Pa.
This pressure acts on piston B of area 1/ 2 m
2
. The total upward force or thrust on B is given by:
Force = Pressure x Area = 100Pa x 1/ 2 m
2
=50N.
A force of 1N thus produces a force of 50N which can be used to lift heavier load.
CONTRACTION AND EXPANSION OF SOLIDS:
Expansion is an increase in volume or size of an object.
Solids expand upon being heated. This is due to increased motion of particles which force each other a
little farther apart according to the Kinetic theory.
Contraction is a decrease in volume of an object. When an object is cooled, the particles or molecules
come closer together.
APPLICATION OF CONTRACTION AND EXPANSION OF SOLIDS:
1. For separating materials. The expansion of solids can be applied to unscrew the bottle top which
is too tight by placing the bottle top in hot water for a moment. The top expands before the heat
reaches the bottle, which makes it a looser fit.
2. For joining materials.
a).Shrink-fitting:
Both expansion and contraction can be used to make a tight fit between tow pieces of metal. For
instance, this can be used to fix the axle into a wheel. The axle can be shrunk by cooling in liquid
nitrogen at -196
0
C until the gear wheel can be slipped on to it. On regaining normal temperature the
axle expands to give a very tight fit.
b Riveting metal plates:
Contraction is applied. In ship –building , steel plates are riveted. A white-hot rivet is placed in the
rivet hole and its end hammered flat. On cooling, it contracts and pulls the plates together.
3. Bimetal Thermostats.
Both contraction and expansion can be applied in Bimetallic Thermostats.
A thermostat is a device which controls temperature of an appliance or room.
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Thermostats are fitted to refrigerators . electric irons, immersion heaters, ovens and to some room
heaters.
4. Bimetallic Thermometer
It contains a bi-metal strip in form of a long spiral. The center of the spiral is attached to a pointer.
The other end is fixed. When the temperature rises, the strip coils itself into an even tighter spiral and
the pointer moves across the scale
Exercise
EXERCISE
2003
Question 4
a. (i) Explain the term absolute temperature?
(ii) convert 25°C to Kelvin scale.
b. Figure 3 shows a gas tube fitted with a piston
i. What will happens to the volume and pressure of the gas when the piston is pressed down
the glasstube?
ii. Explain your answer to 4.b.i
iii. A container with the cross-section of 3m
2
is filled with 9cm
3
of water. Calculate the
pressure on the bottom of the container. If the density of water is 1g/cm
3
iv. Give any two uses of liquid pressure in every day life.
c. In terms of kinetic theory of matter explain why increasing temperature cause the candle melt.
2004
Question 1
a. Table 3 shows the results in an experiment to verify a gas law
Volume (cm
3
)
10
12
14
16
18
Pressure (kpa)
200
169
144
127
114
i. Plot the graph to show the relationship between pressure and volume
ii. What relationship is being demonstrates by this graph
iii. Which variable would kept constant in this investigation
b. (i). A pressureof 50000 pa is exerted by the column of water at the base of the container.
Calculate the height of water.
(ii) Exp0lain why the base of the dam is thicker than the top
(iii) Two identical were filled with liquid mercury with a density of 13.6g/cm3) and water with the
density of 1g/cm3. Which container base feels greater pressure? Explain your answer.
c. Figure 1 is an instrument used to measure gas pressure.
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i. Name the instrument
ii. Read the pressure difference in mmHg
iii. What is the pressure of the gas supply if the atmospheric pressure is 755mm and the
pressure difference is 30mm
2005
Question 1
a. (i) define the term “diffusion”
(ii) In which sate of matter does diffusion occurs quickly?
(iii) Give a reason for you answer
b. Figure 1 is a diagram of a sealed glass tube containing two cotton. One is soaked in
concentrated ammonium and the other soaked in concentrated hydrochloricacid. Gases from the
solutions along the tube and white cloud where they meet.
i. Which of the gases light
ii. Give reason for your answer to b.i
iii. When the experiment is done on the sunny day, it takes a shorter time for the white
cloud to be formed. Give a reason why it happen like that.
c. The volume of the air container is 6cm
3
and has the pressure of 4 atm when te temperature is
27°C Calculate the pressure if the volume is 3cm
3
and the temperature if 177°C
2006
Question 3 e
e. At atmospheric pressure, mercury barometer reads 0.76 m. If one atmospheric pressure is
equal to 10100pa. Calculate the density of mercury.
f. Figure 2 is a diagram showing a hydraulic system being used to raise a load
i. Calculate the pressure piston A exert on liquid
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ii. How much pressure does the liquid exert on piston B
2007
Question 6
a. Given any two properties of liquid pressure
b. Figure 5 is a diagram of as bottle containing water
Calculate the pressure of water on the bottom of the bottle . (Density of water 1g/cm
2
)
c. Mention the term absolute temperature
d. Convert 546K to degree Celsius
2008
Question 8
a. Explain why the candle wax melt when it is heated
b. With an aid of a diagram describe how manometer works’
2009
Question 6.
d. (i) define Gas pressure
ii. Give any three uses of gas pressure
iv. Why does the gas pressure in closed container increases when the temperature
increases?
e. Table 2 shows a results of an experiment in that was done to demonstrate a gas law
Pressure (pa)
0.9
1.1
1.3.
1.5
Temperature (°C)
0
50
100
150
i. Plot the graph oif pressure against temperature
ii. Use the graph to find the pressure of the gas when the temperature was 120°C.
2010
Question 1
a. (i) What liquids is used in hydraulic machine
(ii) Give three examples of hydraulic machine
b. (i) Define absolute zero
(ii) Agas occupies a volume of 200cm
3
at 273K. calculate the temperature of the gas in Kelvin
if the volume increases to 300cm
3
at constant pressure.
2011
Question 6
a. State any two factors that affect gas pressure
b. Explain why the walls of the dam is built thicker at the bottom
c. A rectangular container whose base is 9cm
2
is filled of water with a volume of 27cm
3
. Calculate
the pressure exerted by the water at bottom of the container in N/cm
2
( acceleration due to
gravity = 10m/s/s)
2012
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Question 1
a. State the kinetic theory of matter
b. Figure 1 is a diagram of an apparatus used to measure the pressure exerted by gas
Calculate the pressure exerted by the gas if the atmospheric pressure is 765mmhg
8c. In term of kinetic theory of mater explain why ice melt when put in the sun
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ELEMENT AND CHEMICAL BONDING
STRUCTURE OF AN ATOM
Atom is the simplest particle that made matter
It consist of electrons , protons and neutrons
Electrons
They are negatively charged particle that revolve around the nucleus in their orbital’s.
It is located in energy levels and exist in the lowest energy levels\
They have negligible mass and the move from one energy level when given extra energy
Protons
They are positively charge particle that are found in nucleus.
They have the mass of 1 amu ( atomic mass unit)
Neutrons
They are located in nucleus and has a mass of 1 amu
They have no charge
In a neutral atom the number of electron is equal to the number of protons
Atomic Number ( Z) .
It is the number of protons in an atom
In an neutral atom the number of proton is equal to the number of electron
Atomic mass number (N)
The sum of proton and neutron in an atom is called the mass number ( N)
Example : An element X has the atomic mass number of 31. It has also 16 neutron . Find the number
of protons and electron in the atom
Number of proton plus the number of neutron is equal to the atomic mass number
31 = 16 + Z
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z = 15
Therefore number of electron is equal to 15
A negatively charge particle has charge of negative 3 and has 17 protons. Find the number of electron
in the atom
Since it is a negatively charged particle it means it has gain some electron. Therefore to find the
number of electron in the particle add 3 on the number of proton
The answer is 14 electrons
NB: If it is a cation add the number of charge from the number of proton for anion subtract the number
of the charge on the number of protons
Why atom of different element differ
Atom differ because of the following factors
1. Different atoms has different masses
2. Different atom has different changes on their nuclei and different number of electrons
3. Different atom has different relative atomic mass (RAM)
THE PERIODIC TABLE
During the building of the periodic table two things are looked at
1. One quantity should keep on increasing
2. The other quantity should keep on repeating itself
Electron configuration
It is the arrangement of electron in energy levels
Energy level
Number of electron to fill energy level
1
2
2
8
3
8
4
18
5
32
How to place electron in energy levels
1. Lowest energy level or hells are filed first ie nearest to the nucleus
2. Energy level are filled depending on the number of electron it hold
3. Energy level can be partly or completely filled by filling gradually one by one
Example
Element
Atomic number
Electron
configuration
Hydrogen
1
1
Helium
2
2
Lithium
3
2,1
Beryllium
4
2,2
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Boron
5
2,3
Carbon
6
2, 4
Nitrogen
7
2, 5
Oxygen
8
2,6
Fluorine
9
2,7
Neon
10
2,8
Sodium
11
2,8,1
Magnesium
12
2,8,2
Aluminum
13
2,8, 3
Silicon
14
2,8 ,4
Phosphorus
15
2, 8, 5
Sulphur
16
2, 8 ,6
Chlorine
17
2, 8 , 7
Argon
18
2, 8, 8
Potassium
19
2,8,8,1
Example
Draw the electron configuration for element with 14 protons
The electron configuration for a certain element is shown below
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a). Write its electron configuration
b). How many proton are there in the element
c). What could be possible name for the element
Solution
a). Number of electrons = 18 + 2 = 20
b). Number of protons = 18 + 2 = 20
c). The possible name is calcium
How a periodic table is built
It is built by placing elements with similar electron configuration in the same column and placing
element with increasing electron configuration.
H(1)
He(2)
Li(2,1)
Be( 2,2)
B (2,3)
C (2,4)
N (2,5)
O (2,6)
F (2,7)
Ne (2,8)
Na (2,8,1)
Mg (2,8,2
Al (2,8,3)
Si (2,8,4)
P (2,8,5)
S (2,8,,6)
Cl (2,8,7)
Ar (2,8,8)
K (2,8,8,1)
Ca (2,8,8,2)
Br (2,8,8,7)
Kr (2,8,8,8)
The vertical columns are called Group or families and are made by electron with similar electron
configuration
The horizontal columns or rows are called period or energy levels
GROUPS IN THE PERIODIC TABLE AND THEIR PROPERTIES
Group one elements
They have 1 electron in their out most shell. They are also called alkali metal because they react with
water to form alkaline solution
Examples of group one element include.
Lithium 2,1
Sodium 2,8,1
Potassium 2,8,8,1
Physical Properties of Group 1 Elements
1. They are soft and can easily cut by a knife
2. They are ductile, how easy a metal can be draw into a wire
3. They are malleable, how easy metal can be hammered into any shape
4. They are shiny when freshly cut
5. They have low density and float on water
6. They are good conductor of heat and electricity
CHEMICAL PROPETIES
With air
They react with oxygen to produce a white oxide of similar molecular formula
4Na (s) + O
2
2Na
2
O (s)
4Li(s) + O
2
2Li
2
O (s)
4K (s) + O
2
2K
2
O (s)
With Halogen
They react with group 7 E.g chlorine to halides salts. To initiate the reaction alkali metal should be
heat first. The reaction should be done in a fume board always because it produces a white smoke
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which bis poisonous
2Na (s) + Cl
2
2Nacl (s)
2Li (s) + Cl
2
2Licl (s)
2K(s) + Cl
2
2Kcl (s)
With water
When apiece of alkali metal is plugged into a beaker containing water. It react violently with water
producing a hissing sound .
2Na (s) + H
2
O 2Na OH + H
2
(s)
2LI (s) + H
2
O 2Li OH + H
2
(s)
2K (s) + H
2
O 2K OH + H
2
(s)
Similarities of group 1 elements
1. They are soft and can easily cut by a knife
2. They are ductile, how easy a metal can be draw into a wire
3. They are malleable, how easy metal can be hammered into any shape
4. They show silvery lustre when fleshy cut
5. They have low density and float on water
6. They are good conductor of heat and electricity
7. All react with water to produce alkaline solution
8. All react with air to produce oxide
9. All react with chlorine or halogen to produce a halide salt
NB all group one elements has the same properties because they have one electron in their out-most
shell that take part in chemical reaction and tend to loose that electron
Difference between group one elements
1. Metal become softer to cut down the group
2. Metals burn easily in air down the group
3. They react easily with halogen or chlorine down the group
4. They react easily with water
Why the difference
The reactivity increases because the number of shells increases down the group which makes the
electron to move far from the nuclei and experience very little nucleus attraction. Since electrons are
held loosely they form weak metallic bond and can be easily broken.
GROUP 7 – THE HALOGEN
They are found on the left side of the period table
Examples of Group 7 element include
Hydrogen 1
Fluorine 2, 7
Chlorine 2, 8,7
Bromine 2,8,8,7
Physical Properties
1. They are typical nonmetals and are typical molecular substance
2. They exist as diatomic substances
3. They have low melting and boiling points. Chlorine and Fluorine are gases at room temperature
This Pamphlet Belongs To_ Page 22
while Bromine is liquid and Iodine is a solid
4. They are slightly soluble in water
5. They have choking smell
6. The form molecular compound with other nonmetals for example Hydrochloric acid
CHEMICAL PROPERTIES
They are the most reactive Nonmetals. Halogen tends to gain electron. Reactivity increase up the group
because the smaller halogen has strong nuclear charge that attract electron fast compared with halogen
that have large atom. Fluorine is the most reactive non metal
With group1
They react violently with sodium and other group one and the reaction is exothermic
2K(s) + Cl
2
2Kcl (s)
2K(s) +Br
2
2KBr (s)
2K(s) +I
2
2KI (s)
2K(s) + F
2
2KFl (s)
With other metals
Powdered aluminum and iodine
2Al(s) + 3I
2
2AlI
3
(s)
With other non metal
React with white phosphorus and catches fire by itself E.g in chlorine giving a mixture of products
2P(s) + 3Cl
2
2 PCl
5
(s) Phosphoruspentachloride
and 2P(s) + 3Cl
2
2 PCl
3
(s) Phosphorustrixhloride
With hydrogen
They react with hydrogen to produce hydrogen halide
H
2
+ F
2
2HF explode when mixing
H
2
+ Cl
2
2HCl
H
2
+ Br
2
2HBr
H
2
+ I
2
2HI
TEST FOR HALIDES
Silver nitrate is used to test presence of of halides. Silver nitrate produces precipitate with different
colours
Cl
-
(aq) + Ag
+
(aq) Ag Cl (s) white
Br
-
(aq) + Ag
+
(aq) Ag F (s) pale yellow
I
-
(aq) + Ag
+
(aq) Ag I (s) yellow
PREPARATION OF HALIDES
They prepared by reacting a halide salt with manganese dioxide and concentrated sulphuric acid. . The
halide salt should be sodium or potassium
2NaCl + MnO
2
+ H
2
SO
4
Na
2
SO
4
+ 2H
2
O + MnSO
4
+ Cl
2
2NaCl + MnO2 + H2SO4 Na2SO4 + 2H2O + MnSO4 + Cl2
2NaBr + MnO
2
+ H
2
SO
4
Na
2
SO
4
+ 2H
2
O + MnSO
4
+Br
2
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2NaF + MnO
2
+ H
2
SO
4
Na
2
SO
4
+ 2H
2
O + MnSO
4
+ F
2
2NaF + MnO
2
+ H
2
SO
4
Na
2
SO
4
+ 2H
2
O + MnSO
4
+ F
2
USES OF HALOGENS
Halogen has several uses as follows
Fluorine
It is a pale yellow gas and has the following uses
1. Used in rocket propulsion
2. Used in making uranium fluoride to separate uranium 238 and 235
3. Used as fluro carbon for refrigerant, aerosol propellants anesthetic and fire extinguishers
4. Used in drinking water and toothpaste to reduce tooth-decay
Chlorine
1. Used in manufacturing of PVC pipe
2. Used to kill bacteria and viruses in drinking water
3. Used as bleach in laundries
4. Used as sterilizing fluid in dilute solution
Bromine
1. Used in making disinfectants
2. Used as a medicine
3. Used as fire retardants
Iodine
1. Used as medicine
2. Used as disinfectants
3. Used as photography chemical
GROUP 8 – INERT GASES OR NOBLE GASES
They are gases and exist as mono atomic substance.
Example
Helium 2
Neon 2,8
Argon 2,8,8
They are nonreactive substance because they out shell is completely filled and can not share or transfer
electron to other compounds
They have low melting and boiling points
Although they are nonreactive they have important uses
Examples
Argon
1. Used in bulb to prevent tungsten to react with air
2. Used to provide inertness in arc welding in production of titanium metal
Neon
1. Used in advertising sign it glow red when electricity is passed
2. Used in neon- helium gas laser for operation
3. Used in Geiger Muller tube to detect radioactivity
Helium
1. Provide an inert atmosphere for welding
This Pamphlet Belongs To_ Page 24
2. As coolant in nuclear reactor
3. With 20% oxygen as breathing gas used by deep see divers
4. To inflate tire of big tires of aircrafts
5. Low temperature research
NB
1. Atomic radius. It is distance from the center of the nucleus to the out most shell of an atom
2. Bonding radius it is the distance between to bonded atom nuclei
SOME NOTICED CHANGES IN THE PERIODIC TABLE
Observation 1
The covalent radius increases a from one period to another. For example as we move from lithium to
saodium the atomic radius increase the same as in sodium to potassium.
Explanation to the observation
This so because as we move from one period to another the number of shells increases and the effective
nuclear charge decreases and electron are loosely held making them to have large atoms . The nuclear
charge decreases due increase in shielding effect of electron due to increase in number of shells
Observation 2
Moving along the period the covalent and atomic radius decrease for example moving from lithium to
fluorine.
Explanation to observation
The decrease in atomic radius is so because of the increase in number of protons which pull electron
towards the nucleus making the toms to become small. Increase in atomic number increases the
effective nuclear charge like fluorine has greater effective nuclear charge of +7 while lithium has an
effective nuclear charge of 1. This increase in effective nuclear charge pull electron more strongly
towards the nucleus making the atoms to become smaller that atomic with small effective nuclear
charge.
NB: Fluorineis the most reactive nonmetal and has highest negativity because it has smallest atom and
with effective nuclear charge. This so because as we move along group -7 reactivity increase as move
up the group and also the effective nuclear charge increase as we move from left to right.
COMPOUND FORMATION
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The compounds are formed depending on what the elements do with their electrons
Valences in element
Valence is the number of bond an atom can form with other atoms.
Valence electron is the number of electron that can take part in chemical bonding and is always equal
to the number of electrons in the out most shell
Chemical bond are formed either by transferring electron or sharing electrons
Group
1
2
3
4
5
6
7
8
Atom
Li
Be
B
C
N
O
F
Ne
Valence
1
2
3
4
3
2
1
0
Metal tend to loose electron while gain electron from metals and difficult for them to loose
The number of electron to be lost or gain is equal to the valence electrons
For example the valence electron for Al which is group 3 is 3 while Neon which is group 8 is 0
Metalloid or group 4 elements tend to share the electron because it is difficult for them to gain or loose
electrons.
If non metal form a compound with another nonmetal they share there electrons
Example of compound formation
Al Cl
3 1
Al Cl
1 3
The compound formed is AlCl
3
CHEMICAL BONDING
Bond is the force of attraction that exist between two adjacent atoms
Why elements form bond
Bonds are formed because reactants want to:
1. Want to be at lower potential energy
2. Having a more stable situation
3. To experience more attractive forces
4. To fill the octet or to complete their valence.
Types of chemical bonding
Metallic bonds
This is a force of attraction between fixed metal ions in a piece of metal and free delocalized electrons
that move and surround them atoms.
How a metallic bond are formed
When a metal is gaseous state, electrons are found inside the electron shells of specific atom and are
known as localized electron
When gaseous metal is cooled the atom come closer to each other electrons are attracted by the
different atom nucleus which make electron not to be found in their specific electron shell and are said
This Pamphlet Belongs To_ Page 26
to delocalise.
Atoms loose their electron and become positively charges which attract free electron forming bonds
Characteristics of metals
1. The y are good conductor of heat and electricity
2. They are ductile and malleable
3. They have high Melting points and Boiling points
4. They are strong
5. They are insoluble in water
Ionic Bond
It is the electrical attraction between oppositely charged ions.
How it is formed
It formed when electron are transferred from one atom to another creating two different charge which
later attract one another.
Examples of ionic bonds
Nacl
AlCl
3
or
Characteristics of ionic compound
1. They are made up of ion not atoms
2. They conduct electricity in molten state or in solution form
3. They have high Melting point and Boiling point
4. They are brittle at room temperature
5. They are soluble in water
Covalent bonds
It is the force of attraction that is exists because of sharing of electrons by two adjacent nonmetallic
atoms. Each adjacent atom donates an electron to form a pair that will be shared among them.
Covalent bond are formed when two adjacent atom share electron.
Bonding between hydrogen and chlorine
H- Cl
Metallic bond are formed when gaseous atoms allow their electron to be
delocalised on reaching solid state and feel force of attraction from different
metal ions
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Double covalent bonds
It is formed when two pairs of electron are shared
Each atom donate a pair of electron
For example CO
2
C = C
O = C = O
Triple bonds
Three pair of electrons are needed. Each electron donate three electron to form a bond
Example N
2
N = N
or
Co-ordinate or dative bonds
It the covalent bond which is formed when pair of shared electron is provided by one of the bonded
atom.
The atom that provide electron is called a donor and that accept electron is acceptor. An arrow is used
to represent dative bond.
Example lone pair from nitrogen that make the dative bond NH
4
+
or
Characteristics of covalent compounds
1. They poor conductor of heat and electricity
2. They have low Melting point and Boiling points
3. They low solubility in water
Polar and non polar bonds
Halogen has high electronegative.
Electronegative is the ability of an element to pull electron towards itself when it has been bonded to
another element.
Elements such as fluorine, chlorine iodine, bromine and oxygen tend to pull electron towards them self
when other have form a bond with other elements.
The electrical negativity increases up group seven element because of the effective nuclear charge
increases up the group.
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This make the molecule to have to different partial charge end. One end partially negative while the
other partially positive.
This unequal sharing of electron is referred as polarization and the bond formed is between ionic and
covalent is called a polar bond
Example of polar bond
δ + H – Cl δ-
SULPHUR S
It belong to group 6 .
Sulphur molecule
Sulphur molecules has eight atoms arrange in S shaped ring
Occurrence:
1. It occurs in nature as a free element or combined with other elements in sulfides and sulfates.
2. It can be found in places where volcanoes were formerly active.
Note : Sulphur can be mined in the ground and purified.
Generally , Sulphur is mined or extracted using the Frasch process, named after its inventor
Hermann Frasch. In the process, superheated water at 170
0
C and hot compressed air are forced
underground through pipes-forcing water and molten Sulphur to the surface.
Allotropic Forms of Sulphur:
It can exist in several different solid and liquid allotropic forms just as oxygen can exist in different
forms.
Rhombic Sulphur ( Alpha or Octohedral Sulphur )
It is solid form of Sulphur and is stable at ordinary temperatures below 96
0
C.
It consists of eight- membered rings of Sulphur atoms. .
Crystals of rhombic Sulphur may be prepared by dissolving roll Sulphur in carbon tetrachloride or
xylene and then allowing the solvent to evaporate slowly.
It is the most stable form of sulphur because the S
8
molecules are closely packed and is interlocked
Monoclinic Sulphur Beta or Prismatic Sulphur:
It is stable above 96
0
C
Monoclinic Sulphur crystals have two crystal axes at right angles to each other.
It can be prepared by first melting some Sulphur in crucible at as low temperature as possible. It is then
allowed to cool slowly until a crust begins to form.
Its crystals when cooled below 95
o
C gradually change back into rhombic Sulphur.
Monoclinic sulphur is not stable because the molecules are not closely pack and a change in
temperature less that 95ºC cause it to change to rhombic sulphur
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Lamda Sulphur
It is a liquid allotropic form of Sulphur produced at temperatures just above the melting point of
Sulphur.
It has a straw –yellow colour.
It has almost spherical shape of 8 Sulphur molecules which enable it to roll over easily.
Mu- Sulphur
It can be prepared by heating lambda Sulphur to about 200
o
C It is back liquid.
Note: This temperature. 96C is called the Transition Temperature. Both allotrope ( rhombic and
monoclinic Sulphur) are stable at this temperature.
Amorphous Sulphur:
It is rubber plastic. It is dark brown.
It is made by poring boiling Sulphur into cold water
Uses of Sulphur
1. Used in production of Sulphur dioxide used in production of sulphuric acid
2. Used in vulcanization of rubber
3. Production of chemical used in detergents
4. Manufacturing of gun powder , fire work and matches
5. In preparation of skin ointment and drugs
6. Manufacturing of hydrogen carbonate use in bleaching
7. Used in manufacturing of plastic flowers
8. For making Sulphur concrete in area while acid is mostly used
9. As fungicide ( lead arsenate ) to control blights, mildews and other diseases of plants.
10. Used in making dyes
Physical Properties of Sulphur
a) It is yellow odorless solid.
b) It is insoluble in water.
c) It is twice as dense as water.
d) It is soluble in carbon disulfide and carbon tetrachloride
e) Does not conduct electricity.
Chemical Properties of Sulphur
It reacts with both metals and non metals.
It can react with oxygen.
S ( s ) + O
2
SO
2
( g ).
It can react with magnesium
Mg( s ) + S ( s ) MgS ( s ) magnesium sulfide
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Sulphur dioxide
it is produced by reacting Sulphur with oxygen
S ( s ) + O
2
SO
2
( g ).
physical properties of Sulphur dioxide
1. It is colourless gas with strong choking smell
2. It is heavier than air
3. It is soluble in water
4. It bleaches when is is damp or in solution
Chemical properties
1. It react with water too form sulphuric acid
H
2
O + SO
2
H
2
SO
4
2. It oxidises metal compound and nonmetal compound
E.g aluminum, nitric acid
Uses of Sulphur dioxide
1. Used in bleaching wool , silk, and wood pulp for making paper
2. Used as sterilizing agent in making soft drinks and jam and drying fruit
3. Used in manufacturing of sulphuric acid
PRODUCTION SULPHURIC ACID H
2
SO
4
.
Preparation of Sulphuric Acid.
It is made by either the :
i. Contact process or ii Chamber process
THE CONTACT PROCESS:
Stage 1
Sulphur dioxide is prepared by burning Sulphur or by roasting iron pyrites ( FeS
2
) to produce
Sulphur dioxide
S ( s ) + O
2
SO
2
( g ).
The Sulphur dioxide is purified to remove impurities which might poison the catalyst.
Stage 2
The purified Sulphur dioxide is mixed with air, and passed through heated iron pipes which contain the
catalyst ( Divanadium pentoxide )V
2
O
5
to produce Sulphur trioxide.
SO
2
( g ) + O
2
SO
3
( g ).
Note :
1. The close contact of the Sulphur dioxide and the catalyst gives the contact process its name.
2. Sulphur trioxide( SO
3
) is anhydride of sulphuric acid, and it is an intermediate product in the
manufacture of sulphuric acid. It is solid at room temperature.
Stage 3
The Sulphur trioxide is dissolved first in concentrated sulphuric acid to form oleum ( fuming sulphuric
acid ):
SO
3
(g ) + H
2
SO
4
(l ) H
2
S
2
O
7
( l ) oleum.
Stage four
The oleum is then diluted with the correct amount of water to produce concentrated sulphuric acid.
This Pamphlet Belongs To_ Page 31
H
2
S
2
O
7
( l ) +H
2
O (l ) 2H
2
SO
4
(l ).
Note:
If sulphuric trioxide is added directly to water, sulphuric acid is produced, But ,this is not done on a
large scale because the reaction is too exothermic or violent and boils the acid produced. Thick mist is
produced. The acid mist is very difficult to deal with.
SO
3
(g ) + H
2
O (l ) 2H
2
SO
4
(l ).
Very highly concentrated pure sulphuric acid is produced by the contact process.
The contact process can be summarized as follows
The Chamber Process:
Sulphur dioxide is converted to sulphuric acid by the action of nitrogen dioxide ( NO
2
) and water.
H
2
O + SO
2
+ NO
2
H
2
SO
4
+ NO
Note: This method is used for making sulphuric acid for commercial uses that do not require very pure
or highly concentrated acid E.g for the production of super phosphate fertilizer.
PROPERTIES OF SULPHURIC ACID:
Physical Properties:
1. Concentrated sulphuric acid is a dense oily liquid which is also called oil of Vitriol.
2. Pure sulphuric acid is colourless, though commercial acid may be brown , black or yellow due to
the presence of impurities
3. Concentrated sulphuric acid has a boiling point of 338
0
C.
Note : You must never add water to sulphuric acid , but acid to water. A great deal of heat is evolved
because of the formation of the hydrates:
H
2
SO
4
. H
2
O and H
2
SO
4
.2H
2
O.
Chemical Properties:
1. It react with any metal compound to produce hydrogen chloride gas is produced
2. It reacts with any metal nitrate to produce nitric acid vapour. But the mixture must be heated to
initiate the reaction..
3. Hot concentrated sulphuric acid react with carbon and Sulphur to produce carbon dioxide and
Sulphur dioxide respectively. In each case the acid is reduced to water and Sulphur dioxide .See the
equations below:
C ( s ) + 2 H
2
SO
4
( l ) CO
2
(g ) +2SO
2( g )
+ 2H
2
O( l ).
S (s ) + 2 H
2
SO
4
( l ) 3SO
2( g )
+ 2H
2
O( l ).
4. It reacts with sugar and turns it back. If concentrated sulphuric acid is added to a sample of sugar
the sugar turns yellow, then brown and finally black. The black residue is carbon formed when the
acid has removed hydrogen and oxygen. Similar reactions take place when other carbohydrates are
used.
This Pamphlet Belongs To_ Page 32
5. Sulphuric acid ionizes in dilute water in two stages as follows:
H
2
SO
4
+ H
2
O H
3
O
+
+ HSO
4
-
.
HSO
4
-
+ H
2
O H
3
O
+
+ SO
4
-2
.
USES FO SULPHURIC ACID:
1. It is used in the manufacture of fertilizers such as ammonium sulphate and super phosphate
fertilizers. The acid is used to convert the rock phosphate- calcium phosphate ( Ca
3
(PO
4
) ) into a
more soluble product .i.e. super phosphate.
2. It is used in the preparation of other acids and various sulphates. For instance, Hcl. can be prepared
by treating common salt with sulphuric acid.
3. Removing the surface oxide (pickling ) from metals.
4. Making rayon.
5. Making synthetic soap less detergents.
6. Making drugs, dyes, paints , photographic films.
7. It is used in car batteries.
SULPHATES:
They are salts produced using sulphuric acid.
All sulphates are soluble in water except barium sulphate ( BaSO
4
) and lead sulphate ( PbSO
4
).
Calcium sulphate ( Ca SO
4
) is only slightly soluble in water.
Sulphates are less easily decomposed by heating than nitrates or carbonates.
USES OF SULPHATES:
i. Magnesium sulphate ( Mg SO
4
) is used in medicine as laxative.
ii. Ammonium ( ( NH
4
)
2
SO
4
) sulphate is used as fertilizer.
iii. Barium sulphate ( BaSO
4
) as “Barium metal” is used in diagnostic medical X-ray studies.
iv. Calcium sulphate ( Ca SO
4
.1/2H
2
O ) as “ Plaster of Paris” is used to set bones.
TEST FOR PRESENCE OF SULPHATES:
By a simple test-tube reaction:
- Add few drops of dilutes hydrochloric acid to the unknown. then
- Add few drops of barium chloride.
Result:
A white precipitate of barium chloride is an indicator of presence of a sulphate.
Exercise
2003
Question 3
a. Table 1 shows the first 20 elements of the periodic table
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
i. Work out the atomic number of a Si
ii. Work out the electron configuration of K given that it atomic number is 19
iii. Draw the electron dot and cross diagram of CO
2
iv. How aluminum (Al) attain stability as inert gas configuration
This Pamphlet Belongs To_ Page 33
v. Explain why the melting point of group (VII) elements increases with the increases in
atomic number
Question 6
a. Why are metal good conductors of heat
b. The table 2 shows the atomic number, melting point, boiling point and atomic radii of some
halogen. Use it to answer the question that follows.
Name of
element
Atomic number
Melting
point
Boiling point
Atomic radius (mm)
Fluorine
9
-220
-188
0.07
Chlorine
17
-101
-34
0.10
Bromine
35
-7
59
0.11
Iodine
53
114
184
0.13
i. Which element is a liquid at 25˚C
ii. Why does iodine have biggest radius
iii. Work out the effective nuclear charge of fluorine
iv. Mention any two chemical properties of halogens
c. (i) Give one natural source of sulphur
ii. Mention any two chemical properties of halogen n
6e. State two difference between ionic bonding and covalent bonding
2004
Question 1
Table 1 show the particle in an atom of four elements
Element
Protons
Neutrons
Electrons
Mass
number
Hydrogen(H)
1
1
Carbon ( C )
6
6
12
Nitrogen (N)
7
7
Sodium (
Na)
12
11
i. Copy and complete the table by filling the missing numbers
ii. Which elements in the table will easily form ionic bond
iii. Give a treason for you answer to 1 aii
iv. Work out the molecular formula mass of Methane ( CH
4
)
v. What kind of bonding are involved in methane
vi. Explain your answer to 1 a.v.
b. The dot and cross diagram of calcium chloride is shown below
i. Write the chemical formula of calcium chloride
ii. Explain the meaning of the sign 2+ on Ca atom
c. Table 2 shows elements represented by letter Q , R, L , M, X, Y and Z in the same periodic
table
Group
I
II
III
IV
IV
V
VI I
VIII
Element
Q
R
L
M
X
W
Y
Z
i. Explain the increase of melting point and boiling point from chlorine to iodine
ii. What would happens if the solution of iodine is mixed with solution of potassium
bromide
This Pamphlet Belongs To_ Page 34
4d . Figure 4 is a diagram showing nuclei of two atoms
i. Explain why these atom reacts in the same way
ii. What are atoms of this type called
iii. In what period of periodic table could each belong
iv. Explain your answer to 4d iii
2006
Question 2
a. Halogens such as bromine, chlorine and iodine are prepared by re4acting alkali metal salt with
concentrated sulphuric acid in a presence of catalyst \. Name any salt where each of the
following can be prepared
1. Bromine
2. Chlorine
3. Iodine
b. Sate two properties of halogens
c. Draw the electron shell diagram for a fluorine atom (
19
9
F)
d. i. arrange the
127
53
I ,
35.5
17
Cl and
80
35
Br in order of increasing reactivity
ii. Explain difference in reactivity of elements in 2 d .i
e. State any chemical properties of sulphur
f. Explain with the aid of diagram why rhombic sulphur is more stable that monoclinic sulphur
2007
Question 5
a. Table 1 shows the atomic number and electron configuration of some elements
Element
Atomic mass
Electron
configuration
A
18
2,8,8
B
10
2,8,
C
20
2,8,8,2
D
12
2,8,2
E
2
2
F
9
2,7
i. Identify element that comes from period 2
ii. Which two element can form positive ions
iii. Give a reason for your answer 5a. ii
iv. Give one property of element A
2008
Question 1
a. Element x has a mass of 39 amu and atomic number 19
i. How many protons are in the atom
ii. What would happen if element X react with water
iii. Give a reason for you answer to a. ii. above
b. Magnesium and chlorine can be represented as
24
12
Mg and
35.5
17
Cl respectively
i. What are the valences of magnesium and chlorine
ii. What is the molecular formula of the compound formed as result of magnesium
reacting with chlorine
17 P
20N
17 P
18N
This Pamphlet Belongs To_ Page 35
c. Draw the dot and cross diagram of carbon dioxide ( CO
2
) given that carbon is in group 4 and
oxygen is in group 6 of the periodic table
d. What type of bonding exist in carbon dioxide
e. Give a reason for your answer in 1 d above
f. Table 1 show the melting points of metals, A ,B ,C and D
Table 1
Metal
Melting point
A
240
B
3370
C
1539
D
120
i. Which metal would be suitable for filament of the bulb
ii. Give a reason for your answer to 1.f. i.
g. Sulphuric acid (H
2
SO
4
) is used as hydration agent.
i. Name the products in the hydration of sucrose ( C
11
H
22
O
11
)
ii. Give four uses of sulphuric acid
200 9
Question 1
a. Define the electron configuration
b. Figure 1 is a graph of atomic radius across the period against the number for some element in
the periodic table
i. To which group of the periodic table does element O belong
ii. Give a reason for your answer to bi.
iii. Why there s a sudden increase in atomic radius from F to Na?
iv. In term of atomic radius explain why they is a difference in reactivity between F and Cl
c. (i) .Mention two use of sulphur
ii. Give two physical properties of sulphur
d. Table 1 shows the number of valence electron and valence of some elements
Element
Number of
valence electron
Valence
Li
1
1
Be
2
2
N
5
3
O
6
2
i. How element N can attains stable electron configuration?
ii. Give a reason for your answer
iii. What is the formula of the compound that is formed between Li and O
iv. Give a charge of Be
2010
a). (i) State the three properties of metals
ii. Explain why potassium is more reactive than sodium
b. Figure 1 shows a diagram of atomic number and mass number elements R and Q
i. Write the electron configuration of element R and Q
ii. To which period and group of the periodic table does the element R belong
11P
12n
17 p
18n
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iii. Draw a dot and cross diagram of the compound that would be produced when R react
with Q
c. Explain why helium, which has 2 valence electron is taken as a group 8 element
d. State any two use of sulphates
2011
Question 4
a. State three ways how atoms attains stability
b. Explain how ionic bonds are formed
c. Table 1 shows the atomic number of boiling points of some element represented by letters D,
Q, T , X and Z
Table 1
Element
Atomic
number
Boiling point
D
3
1342
Q
13
2467
T
16
445
X
18
-186
Z
19
760
i. Identify any two letter represent element which belong to period n3 of the periodic table
ii. Which element is in gaseous state at room temperature (25˚C)?
iii. What type of bonding would exist when element Q react with element T
iv. Write down the chemical equation for the reaction that would occur between D and T.
2012
Question 6
a. The table below shows electron configuration of elements R,S T, U and V.
Element
Electron configuration
R
2,7
S
2,8,6
T
2,8,2
U
2,4
V
2
i. Which element in the table belongs to period 2 of the periodic table?
ii. Give a reason for your answer
iii. Give a pair of elements that would form an ionic compound when they react
iv. Draw an electron dot and cross diagram for the compound formed when S combine with
U
b. State any three physical properties of halogens
c. Explain what happens id chlorine is mixed with potassium bromine solution
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FORCE AND MOTION
VECTOR AND SCALARS
SCALARS
Scalars are quantities with magnitude only without direction
Example of scalar quantities
Charge, Energy, Frequency, Length, Mass, Power, Pressure, Speed , Temperature, Time, Volume
Vectors
Vector quantities that have both magnitude and direction
Examples of vector quantities
Acceleration, Displacement, Electric field , Electric current, Force , Magnetic flux density, Moment,
Momentum, Torque, Velocity, Weight
Presentation of vectors
Vectors can be represented using the following ways
1. By using a line with an arrow. The arrow point to direction of the vector and should be proportional
to the magnitude of the vector quantity
5 N 3N 10N 15N
2. By using a bold face letters
A or AB D B
3. By using a letter with an arrow on top of the letters E.g AB
4. Using small letter a
PROPERTIES OF VECTORS
1. Property of equality: Vectors are said to be equal if they are pointing in the same direction and
have the same magnitude
Example 8N
5N 8N 4N 4N
5N
2. Negative property : Two vector are opposite if they are pointing in the opposite direction and have
the same magnitude
9N 7.1 N 7.1 N 15N 15 N
9N
3. Vector can be shifted from one point to another without changing direction
4. Parallel property: Two vectors are parallel if they are pointing in the same direction and are
equidistant from one another
5. Vector can be multiply or divided by the scalar quantity
6. Vectors can be added or subtracted
ADDITION OF VECTORS
Vector lying in the same plane pointing in the same direction or opposite direction can be simply added
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Example1 7N + 5N = 12 N
Example2 15N- 9N = 6N
Note that in this example vector are pointing in opposite direction we need to subtract to get the
resultant vector
Example 3 3N + 10N = 13N
ADDING VECTOR AT AN ANGLE
Using Pythagoras theorem
This method is used to add vector that act at right angle to each other .
The Pythagoras theorem state that the sum of the square of opposite side is equal to the square of the
third side ( the hypotenuse side)
AB
2
+ BC
2
= AC
2
T he direction is given by using the angles deviated from the horizontal
Example.
Find the resultant when the air plane flies due to east and a velocity of 120 per hr and is heading wind
that is traveling at velocity of 90km per hour to the north direction
First draw the diagram
Using Pythagoras
AB
2
+ AC
2
= BC
2
90
2
+ 120
2
= BC
2
8100 + 14400 = BC
2
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BC
2
= √12500
BC = 150 km /hr
To find the direction of the resultant use tan x = 90/120
Tan x = 0.75
Tan
-1
0.75
= 36.9 degrees
Therefore the airplane will travel at a velocity of 150 km/ hour at a direction of 36.9 degree
By scale drawing / graphical method
A Suitable scale is used to calculate the resultant of the vectors acting an angle
Examples
Cyclists traveling due to south 12km from point A and later turn east and travel a distance of 8km. Find
the shortest distance and direction where the cyclist from the starting point.
Let our scale be 1 cm represent km on the a paper
Length of 12 km on a paper is 3cm
Length of 8km on the a paper is 2cm
A
Draw AB as follows
Then draw BC
Join the point A and C
To find the resultant
Measure AC using a ruler and multiply the value with the scale
Measure angle BAC to find the direction of the cyclist
NB this method works better when you are adding vector that are acting at an angle of 90 degrees
Triangle Rule Tip to Tip
This is the method where one vector comes on the head of another vector to from a triangle
Step to follow when doing triangle rule
Step1. First draw the 9 N vectors to a scale using the ruler.
Step2. Shift the 6N vector so that it tail lies on the head of 9N vector and
Step 3. Then draw vector so that the head is on head of the 9N and the tail is on 4N
NB tail head of vector
Parallelogram rule
It state that if two force are acting at a point are represented in the size and direction by the side of
the parallelogram drawn from the point , their resultant is represented in the size of the direction by
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the diagonal of the parallelogram drawn from the point
vector are used as the side of the parallelogram and complete the parallelogram and draw vectors from
the tail to head and measure the length and convert using the scale
Example
Find the resultant of the following vector
First choose the suitable scale for example 1cm to represent 3N
Draw two vectors to a scale using a ruler at an angle of 30˚
Mark off length of each vector
Complete the parallelogram with the adjacent side
Like
Draw the diagonal from point you started
Measure the diagonal
RESOLUTION OF VECTORS
When you are pulling the object at and angle the object move horizontally until it reaches where you
are and later raises up
It shows that the they are two forces that act on the object that is being pulled at angle e.g. the
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horizontal and vertical force. These two forces are called vector components
They are two vector components the horizontal component and vertical component.
In resolution of vector these two components are determined.
HOW TO RESOLVE VECTOR
They are two ways how vector can be resolved.
Graphical method and buy using trigonometric ratios
To find the components use the following formulas
GRAPHICAL METHOD
Parallelogram rule can work in reserve.
The single vector act as the diagonal of the parallelogram. To resolve follow the following steps
1. Draw the given vector to scale which will act as the diagonal
2. Draw two lines that should be perpendicular to each other from the origin of the vector like
3. Draw another two lines so that you complete the diagram to form a parallelogram like this
4. Measure the length of the vertical line to fine the magnitude of the vertical components of the
vector and the horizontal line to fine the magnitude of the horizontal vector component
By using a triangle rule
1. Draw the single vector to act as the hypotenuse of the triangle like follows
2. Draw a line to scale from the origin of the vector. Draw another line to complete the triangle by it
should act at right angle from the end point of the first line to meet the tip of the vector. Like this
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3. Measure the length of the line draw. The vertical represent the vertical component while the
horizontal represent the horizontal component.
Resolving by calculation using the trigonometric Ratios
Trigonometric ratios can be used to find the magnitude of the components of the vector if one angle is
given.
Consider the right angled triangle below
Cos α = Adj/ Hyp
Cos α= AC/ AB
AB Cos α = AC
and
Sin α = OPP/ HYP
Sin α = BC/ AB
AB Sin α = BC
But AB is the resultant vector R , BC is the vertical component and AC is the horizontal component
Therefore
R Sin α = vertical component
R Cos α = Horizontal component
Example
A force of 60N is at an angle of 30 degree to the horizontal . Find the perpendicular vectors
Vertical vector =R sin 30
60 sin30
60 x 0.5
α
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30 N
Horizontal vector = Rcos30
60 x 0.8660
1.1.1.1.51. 9615 N
52N
Example 2
Resolve the following vector into their vectors component
20N
a) b). 45N
Horizontal component = 20 Cos 90 Horizontal components = 45Cos180
= 20 x0 = 45x1
= 0 = 45N
Vertical components = 20sin90 Vertical components = 45Sin180
= 20x1 = 45x0
= 20 = 0
NB. Perpendicular vertical vector has no horizontal components this is a reason why the body just
move upwards when you are pulling at 90 degree angle and the vertical component is equal to the
vertical vector likewise the horizontally lying vector.
LINEAR MOTION
Distance and Displacement
Distance is the actual path length traveled by a body and is scalar quantity while displacement is the
shortest distance or the straight line distance in a particular direction
Speed and Velocity
Speed is the rate of change in distance with time while velocity is the rate of change of displacements
Mathematically speed = distance covered
time taken
s = d/t where d is distance covered and t is time take
Velocity = Displacement
time taken
Example. A millipede moves a distance of 7m in 3,5 s. what its average speed?
Answer .
Speed = d/t
= 7m /3 = 2 m/s
Acceleration
It is the rate change in velocity
Acceleration = change in velocity
time taken
a = ( v -u) where v is the final velocity and u is the initial velocity and t is time taken
t
Example. A cyclist increase her velocity from 8m/s to 14m/s in 3 seconds. Find the acceleration of the
cyclist
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a = (v-u) /t
= (14 – 8)/3 = 6/3 = 2m/s/s
Example 2 ; a car traveling at 30m/s changes its speed suddenly to 5m/s in 5seconds find ite
acceleration?
a = (v-u)/t
= (5-30)/5
= -25/5
= -5m/s
NB: Acceleration is positive if the object increases speed and is negative if the cyclist reduce speed.
Negative acceleration is called decelerationor retardation
MOTION GRAPHS
Motion of object can be represented on the graph. The pattern of the graph can tell the type of the
motion taken by the object
DISTANCE - TIME GRAPHS
Graph of stationery object Graph of object moving at constant speed
Graph acceleration object Graph of deceleration object
Speed from distance time graph
INTERPRETATION OF DISTANCE TIME GRAPH
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Part of the graph
Interpretation
AB
The object move with constant speed
BC
The object was are rest ( stop)
CD
Change direction. It was going in the opposite direction it turn back
From graph B we can see that distance is directly proportional to time taken
Mathematically S t
S = Vt where V
Proportional constant
Slope of graph line
Speed of object
Distance traveled per unit time
SPEED - TIME GRAPH
Graph of object moving at constant speed Graph of object moving at constant acceleration
Graph of object moving at variable Graph of object moving at variable deceleration
Acceleration
Graph of object moving at constant deceleration
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Interpretation from Speed -time graph
Interpretation
Part of the graph
Interpretation
AO
Constant acceleration
AB
Constant velocity , zero acceleration
BC
Constant acceleration
DC
Deceleration
SLOPE OF SPEED- TIME GRAPH
From graph speed(V) is directly proportional to time taken
V = at where a
is the change in speed
Proportional constant
Acceleration of object
Speed per unit time
AREA UNDER SPEED TIME GRAPH
Area of triangle ABC = ½ (base x height) = ½ AC x BC
But from the graph AB = time (t) and AC is speed (V)
Substituting in the formula of area of triangle we get area of triangle = ½ (v x t)
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But speed multiply by time is distance covered
Therefore Distance (S) = ½ (v x t)
S= ½ vt
NOTE: The area under speed-time graph is equal to the distance traveled
Acceleration formula
Total distance = area under the
graph line
but are of trapezium = ½ (sum of
parallel sides x perpendicular
distance )
Area = ½ ( OA+ DB ) x OD
But AO = u , DB = v and OB = t
Then area = ½ ( u+v) t
but area is distance S
Therefore S= ½ (u+v)t
Arranging the equation S/t = (u+t)/2
and S/t is average speed. Therefore average speed is ½ (U+V)
We know that the area of the triangle is ½ (vt)
But v= at
substitute V for at we get ½ at
2
. ….................................................1
But the area of the rectangle is L x B where L is the time taken and B is the initial speed on the object
according to the above graph
Area = ut------------------------------------------------------------2
combining 1 and 2 we get
S = ut +½ at
2
.
IMPORTANT EQUATION FROM SPEED -TIME GRAPH
S= ½ vt
S= ½ (u+v)t
average speed V = (v+t) /2
a = ( v-u)/t
S = ut +½ at
2
.
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Example.
A boy runs from the rest down the slope 50 m long. The time take to reach the eand point was
10second.
a). Calculate the boys average speed
b). What was the boy's speed at the end of 10 seconds
Average speed = total distance /time
V = 50/10 = 5m/s
Average speed at the end of 10s = final speed + initial speed
2
V = v+u
2
but the initial speed is zero
therefore 5 = v+0
2
2x5 = v
v = 10 m/s
Example 2
Below is speed- time graph of motorist
Find total distance covered by the motorist
Divide the graph as above for simple calculation and later add the area of each segment
Area section A is the same as area of rectangle
Distance = 5x 15 = 75m
Distance for segment B = ½ (5+10)5 = 50 m
Distance for segment C = ½ ( 15+ 10) 10 = 75m
Total distance = 75 +75 +50 = 200m
LINEAR MOTION EQUATION
We know that acceleration = v-u
t
a = v-u or v = u +at .................. equation 1
t
Average velocity = displacement
time taken
V = u+v or S/t or S =( v+u)t ...... equation 2
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2 2
If we substitute equation one in into equation two for v
S = [ u + ( u+at)]t
2
= ( u + u+ ut)t
2
= ut + ut + ut
2
2
= 2ut + ut
2
2
= 2ut/2 + ut
2
/2
= ut + ½ ut
2
….................................. equation 3
Rearranging equation one we get t =(v-u) /a
substituting in equation 2
S = (u+v) x(v-u)
2 a
multiplying both side with 2as
2as = (v+u) (v-u)
2as = v
2
- u
2
v
2
= 2as + u
2
Example.
A car starts at rest and reaches the velocity of 30m/s in 5 minutes. Assume that the acceleration was
constant, calculate:
a) The acceleration of the car
b) The distance covered in this time interval
Solution
First list what have been given to know the best equation you can use
u = 0 ( at rest), V = 30m/s , t= 5 minutes, a = ? , S = ?
a). = ( v-u)/t
= ( 30 – 0) /5
= 6m/s/s
b). S = ut +½ at
2
= 0x5+ ½ x6x5
2
= 0+75
= 75m
Example 2. A cyclist moving a speed of 15m/s accelerates at 2m/s/s for 30 seconds. She then traveled
at constant speed for 40 second and finally decelerate uniformly at 4m/s /s and stops. Find the
maximum speed and total distance covered by the cyclist
Solution
There are three stage taken by the cyclist:
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first stage:
u = 15m/s , a = 2m/s/s , t = 30 s s =? v = ?
v= u+ at = 15m/s + 2m/s/s x 30 = 15m/s+ 60m/s = 75m/s
Distance covered in first stage
S = ut +½ at
= 15x 30 + ½ x2x 30 x30
= 450 + 900
= 1350m
Second stage
v= 75m/s t = 40s
S = vt
= 75x 40
= 3000m
Third stage
u= 75m/s , v = 0 a = - 4m/s/s
v
2
= 2as + u
2
s = v
2
+ u
2
2a
= 0 + 75 x75
2 x -4
= 5625 /-8
= 703.125m
Total distance covered = 703.125 + 3000+ 13500
= 5053.125
NEWTON LAWS OF MOTIONS
Newton First Law of Motion
It states that a body will remain at rest or continue to move at constant speed in a straight line if
there is no unbalanced force. Or a body at rest will remain at and a body moving at constant speed
will do so at straight line provide that the they no any external force acting on it
Implication of the first law
a) Force is something that is capable of changing the state of rest or the uniform speed of the body
b) Matter has inbuilt resistance to change its state of motion
c) An external force is needed to change the state of rest / motion of the body
d) For the body to be at rest or at uniform speed the resultant force should be equal to zero
e) If the body has zero acceleration either the body is at rest or at constant speed
The tendency of the body to maintain its state of rest or uniform speed in a straight line is called
inertia
NEWTON SECOND LAW OF MOTION
It is also called the law of acceleration because it describe the motion of body when unbalanced force
act on it
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It states that the rate of change of momentum of a body is proportional to the resultant force or the
rate of change in acceleration of the body resulting from an applied force is proportional to that
force and inversely proportional to the mass of the body
Implication of second law
The law indicates that force can be measured by multiplying the acceleration of the body for the known
mass
F = mass x acceleration
= 1kg x 1/s/s
= 1kgm/s/s
= 1 N
I Newton is the force that cause a body of 1 kg to accelerate 1m/s/s
Application of the second law
example1.
A horizontal force of 20N is applied to the mass of 2kg on frictionless surface. Find the acceleration of
the body
F= ma
a = F /m
a = 20/2kg
= 10m/s/s
A body moving at 20m/s along a straight line is brought to rest after traveling 40 m. find the breaking
force if the mass of the body if 15kg
First find the acceleration of the body
a = u
2
- v
2
2s
a = 0 - (20 x20)
2x 40
=- 400/80
= - 5m/s/s
f= ma
= 15 x 5
= 75N
Momentum and Newton second law
Momentum is
The ability of the body to move ahead or forward
To keep on going
Having enough movement, or push
Factor affecting the momentum of the body
Amount of velocity
Mass of the body
Momentum = mass x velocity
M= m x v
We know that the acceleration = change in velocity
time
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a = (v - u )
t
But from Newton’s second Law
F = ma where m is the mass of the object and a is the acceleration of the body.
Substituting a with (v-u)t in Newton second law we get equation ( F = Ma)
F= m((v - u )
t
F = mv- mu
t
From equation above we can say force is the rate of change in momentum since (Mv – Mu ) is
change in momentum of the body that is moving.
Example
A lorry of mass 20000kg is traveling at 30m/s. The driver apply brakes and the lorry stops after 5 s.
find the average braking force
F = mv- mu
t
= 20000( 0-20)
5
= -80000 N
Rearranging F = mv- mu
t
Ft = mv – mu
From the rearrangement we can say that Ft is rate of change in momentum called the impulse and is
measured in Ns
Application of Newton second law and momentum
1. Crumple zone in cars to safeguard passengers.
2. Seatbelt to reduce time taken by stretching to reduce time of collision
3. Air bag on the steering wheel which protect the driver
NEWTON THIRD LAW OF MOTION
It states that every action there is an equal opposite force of reaction
IMPLICATION OF THE THIRD LAW
1. Force act in pair as a result of the interaction of the bodies
The reaction and action force are equal in magnitude and opposites in direction and they act on
different bodies
The force is created irrespective of motion
From the third law we can say
F
1
= F
2
But F = mv- mu
t
F.t = mv – mu
Ft is known as impulse or rate of change
In momentum since they are equal and oppose force
The impulses of opposite forces should be the same after collision
Law of conservation of momentum
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It state that when two bodies act on each other , their total momentum stays the constant if no out side
force act on it .
Total momentum before collision = total momentum after collision
m
1
v
1
= m
2
v
2
Example 1
A bullet of the mass 10g and traveling at velocity of at velocity of 100m/s hit the wood having a mass
of 1kg. Find the final velocity of wood after collision
Answers
m
1
v
1
= m
2
v
2
10g x100 = 1000 x v
2
1000 1000
v
2
= 1000/1000
v
2
= 1m/s
Example 2
Two bodies A and B are going in opposite direction. A has a mass of 5kg and is traveling at velocity of
3m/s while B has a mass of 3kg with velocity of 2m/s/s. Accidentally the collide to each other and stick
to each other. Find the final velocity after collision
Solution
Velocity 3m/s Velocity 2m/s
Before collision
Momentum of body A = 5kg x 3m/s = 15kgm/s
Momentum of body B = 3kg x 2m/s = 6kgm/s
Total momentum = Momentum of body A – Momentum of body B
Total momentum = 15kgm/s- 6kgm/s
= 9kgm/s
After collision
Total mass of the body = 8kg
Total momentum before collision = Total momentum after collision
Mv = 9kgm/s
8kg x V = 9
v = 9kgm/s
8kg
v = 1. 125m/s
Application of the third law
The jet and jet engine are designed I such way that it produces the gas which travels backward pushing
the rocket forward
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Falling bodies
When an object is falling there are two force act on the object
Air resistance
Weight
Air resistance or up thrust force pushes the object upward while weight pushes the object downwards
At first they are two unbalanced force, the object accelerate. The weight of the object is greater than the
upward force, up thrust. W eight > Up thrust
At sometime the two force balance each other upward force is equal to down force (weight = up thrust
force) . At this point the object move at constant speed and it is calledTerminal Velocityuntil it lands
on the ground
Factor that affect speed of free objects
1. Size of object: Large object has small terminal velocity because the air resistance s always greater
that the it mass. Small object has high terminal velocity because their surface area is small which
make air resistance or retardation force to be small
2. Weight of object: Light object experience high resistance and has very small terminal velocity
compared to heavy objects which accelerates longer distance
3. Medium: Some medium has little resistance eg in vacuum and air. While liquid with large
molecules has high resistance which make the object move very slowly
Falling in liquids
From the surface of liquid (from point A) the object will accelerate until it reaches B . the object will
accelerate because there is unbalanced forces acting on the object.
The weight of the body which is acting downward is greater that the upward force which is exerted
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friction of the liquid and the up thrust.
Up thrust + friction force < Weight of the ball
U+ Fr < W
From point B to the bottom of the container the object will move with the constant speed until it
reach the base of the container. During this period the weight is equal to the sum of the upthrust force
and the friction force exerted by the atoms of the liquid.
Like
Upthrust + friction force = weight
U + Fr = W
NB the following factor affect the motion of the object in liquids
1. Viscosity, liquids which are viscous exert large friction force compared to liquids which are not
viscous. In viscous liquid the object get its constant velocity very late while in
2. Initial speed of the object in the liquid
3. Size of the object
Free falls
It is the falling of the object due to force of gravity without any external influence such as up
thrust , weight air resistance
Falling in vacuum
If a feather and steel ball is through to pass through the vacuum.
The speed will be the same because the upthrust force and friction force is negligible compared in air.
No friction force will be exerted on the two body hence they will move on the same and reach the
constant speed once they are in vacuum space
While if a fetcher and a steel ball fall in container with air the steel ball will land fast that the feather.
This so because the feather will experience more fiction force or the upward force is greater that the
downward force causing the feather to move very slowly.
There is great downward force acting on the steel ball that the upward force hence it will land fast than
the feather.
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Falling in different liquids
To show that different liquids has different resistivity or to show that media affects the speed of the
object falling in liquids set you apparatus as follows
Using transparent jars of the same size and identical ball bearings throw the two ball bearing at the
same time and observe which one will land first.
The one that will land first has low resistance while the one will land last has high resistance.
On the above experiment oil has high resistance therefore it will land last and water it will land first.
Falling with parachute
Once the parachutist has jumped from the airplane, with the parachute unopened.
The parachutist accelerates because they are two unbalanced forces acting on her.
Like up thrust + friction < weight
U + Fr < W
Later get her constant velocity because the two force are balanced like
U + Fr = W
When the-parachute is opened she decelerates because the are two unequal balance force. The
upward force is greater than weight because the parachute trap air that will reduce her speed.
Like U + Fr > W
This makes the parachutist to decelerate
Later the parachutist will gain a new constant velocity lower that the first one which make him to land
safety
Graph of falling with a parachute
Between A and B the paratuchist accelerate because there is tow un balanced forces , weight is
greater that up thrust force since the parachute is un open .
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Between B and C the parachutist gain a constant speed since thre is two balance force acting, the
weight is equal to up thrust force.
Between C and D the parachutist decelerate when the parachute is open the upthrust force is greater
than weight .
Between D and E the parachutist gain constant speed until landing since upthrust is equal to weight
Exercise
2003
1. b. Figure2 is distance – time graph for cyclist. Use it to answer the question that follows
i. What is the total distance travelled by the cyclist
ii. How long it will take the cyclist to cover the distance?
iii. Describe the motion of the cyclist from 20Pm to 4:30 pm?
iv. Calculate the average speed of the cyclist during the first 2hours of the journey
c. state whether “distance “ is a scalar or vector quantity . give a reason for your answer.
2004
Question 8
a. Explain how you can find the speed of an athlete
b. State two sources of errors in 8a.
2005
Question 4
a. Forces of 40N and 50N are acting at right angles. Draw to scale the diagram to find out the
resultant force. (Use 1 cm to represent 10N)
b. Figure 3 is a diagram showing a set up in an experiment that was carried out to investigate how
a feather and a metal would fall in a vacuum
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i. Name the resultant force acting on the ball
ii. State the direction of the resultant force
iii. If the feather and the ball wre allowed to fall at the same time. Draw the diagram to
show the position
iv. Explain you answer to b. ii. Above.
2006
Question 6
a. Define acceleration
b. The speed of runner dropped from 80m/s to 60m/s in 4 second
i. Calculate the average deceleration of the runner
ii. If the runner maintained the deceleration in 6b.i. After how long did the speed reach
zero
Question 8.
a. A metal ball released on the surface of lubricating oil in a tall glass tube. Explain changes in
velocity as it falls through the bottom of the tube
b. Derive the formula that that pressure of the liquid depend on its density and depth
Question 4 PPII
Explain how you can find the density of cooking oil given empty clear and water
2007
a. Figure b3 shows a sped time graph of car in motion
i. Describe the motion of the car from point A to D
ii. Calculate the distance covered when the car moves from point A to B
b. Figure 4 shows the resultant of two forces
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i. Complete the diagram to show the vertical and horizontal components
ii. Calculate magnitude of the horizontal components
2008
Question 6
a. State two factors that affects the terminal velocity of free falling in air
b. Figure 6 is a diagram showing an object falling at terminal velocity. F
1
and F
2
are forces
acting on the object
i. Name the force F
1
and F
2
ii. What would be the relationship between magnitude of F
1
and F
2
at terminal velocity
iii. State the Newton second law of Motion.
c. An object of mass 2000kg accelerate uniformly from rest to a velocity 20 m/s in 4 seconds .
calculate
i. Acceleration of the object
ii. Force required to produce the acceleration in 6c.i.
Question 4 PPII
a. Describe and experiment that could be carried to determine the average speed of a an athlete
given the following materials: tape measure, stop watch and a whistle
b. State two source of errors
2009
Question 6
a. Define resultant vector
b. Why the speed a scalar quantity while velocity a vector quantity?
c. Two forces of magnitude 240N and 240N are being used to pull a boat at and angle of 60˚ to
each other. Find the resultant force using a scale diagram ( use 1 cm to represent 100N)
2010
Question 3
a. Mention any two examples of scalar quantities
b. Explain how vectors are represented on a paper
c. Figure 3 is diagram showing rowing of a boat on a lake
12N
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i. Explain how rowing cause the boat to move
ii. Which Newton’s law of motion is demonstrated in this diagram?
2011
a. What is meant by free fall of an object
b. State forces acting on the object in air
c. Figure 1 is a velocity time graph showing a journey made by a cyclist
Calculate the average sped of the cyclist
2012
Question 7
a. State Newton’s third law of motion
b. Give two properties of vector quantities
c. Mention one method of adding vectors acting at angles to each other.
d. A car decorates at the rate of 3m/s/s for 5 minutes. If the initial speed is 20m/s , calculate the
final speed
ORGANIC CHEMISTRY
This is the study of chemistry of the compound that are finding in living organism. This include all
compounds with carbon atoms in their molecular structure excluding simple ones like carbon dioxide, carbon
monoxides and carbonates
All compound that contain carbon atom are known as organic compounds
Why carbon form many compounds
Carbon has valence of four and can therefore form four bonds. Carbon atom can join up with other carbon
atoms to form long chains of carbon atom inside molecules of different compounds. It form skeleton where
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other atoms are attached.
TYPES / CLASSES OF ORGANIC COMPOUNDS
They are categorised depending of the type of atoms attached to it. There are three classes and these are
Hydrocarbons
They contain carbon and hydrogen atoms in their molecules . Their main source petroleum. Hydrocarbon are
obtained from petroleum by fractional distillation
Types of hydrocarbons
They are two types of hydrocarbon
Aliphatic hydrocarbons
Are hydrocarbon that do not contain a benzene ring in their molecular structure
Examples include Alkanes, Alkene and alkynes
Aromatic hydrocarbons
They contain benzene ring in their molecular structure
Examples is arenes ,
Oxycarbons:
They contain oxygen in their molecular structure Examples include ,alkanols, aldehyde, ketone , carboxylic
acids
Nitro carbons
They contain nitrogen in there structures. Example include amines
HOMOLOGOUS SERIES
It is the collection of organic compound with similar properties and related formulae and structures.
PROPERTIES OF HOMOLOGOUS SERIES
All homologous series has the following properties
1. Functional group: the determine the physical and chemical properties of members of the homologous
series
2. General formula: it is used to determine the formula of different members in the homologous series
3. Common nomenclature: this is the system of naming of homologous members
It use the following prefix
Number of
carbon atoms
Prefix
1
Meth-
2
Eth-
3
Prop-
4
But-
5
Pent-
6
Hex-
7
Hept-
8
Oct-
9
Non-
10
Dec-
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4. Difference among successful members: in all homologous series they have a difference of -CH2-
between successful members
5. Increase in size: moving along each homologous series, size and masses of the molecular increases,
strength of IMF increases
6. Chemical properties: all members have the same chemical properties.
ALKANES
It belongs to hydrocarbon and contains hydrogen and carbon atoms only
It has no functional group and contains only single bond of C-H and C- C only. This makes Alkanes not to be
reactive or to be inert because they have strong intermolecular forces and strong covalent bonds
The general formula for Alkanes is C
N
H
2N+2
. Where N is the number of carbon atoms. This is used to
calculate formulae of Alkanes
Examples: Find the molecular formula for Alkane with 5 carbons
C
N
H
2N+2
substituting N for 5 it becomes C
5
H
( 5X2) +2
then the formula is C
5
H
12
System of naming
It has the same prefix with other homologous but the suffix is -ane
No Of C
Formula
Structure formulae
Name
1
C
1
H
4
Methane
2
C
2
H
6
Ethane
3
C
3
H
8
Propane
4
C
4
H
10
Butane
5
C
5
H
12
Pentane
6
C
6
H
14
Hexane
7
C
7
H
16
Heptane
8
C
8
H
18
Octane
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9
C
9
H
20
Nonane
10
C10H22
Decane
CYCLOALKANES
They have the general formulaCNH2N. The carbon atoms form a ring. The system of naming is the same
but cyclo is added to corresponding Alkanes name e.g. cyclo-propene
Name
Structural formula
Molecular formula
Symbol
Cyclopropane
C
3
H
6
Cyclobutane
C
4
H
8
Cyclopentane
C
5
H
10
Cyclohexane
C
6
H
12
ISOMERISM OF ALKANES
Conformations
Conformations are compounds with the same molecular formula and structure formula
Examples of conformation of pentane
NB Conformation has different shapes but when stretched the form the same shape
Isomers
Isomers are compounds with the same molecular formula but with different molecular structures, examples
of pentane isomers
RULES TO BE FOLLOWED WHEN NAMING ISOMERS
1. Locate or number the carbon chain with longest carbon atoms or the parent chain. Numbering
should start to nearest first junction in the parent chain.
Example
2. Count the number of the carbon in the parent chain to give the basic name for example
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I has six carbon the basic name is hexane
3. Name any group attached to the parent chain using the following prefix and state the number on
the junction
Prefix
Structure
Methyl
-CH
3
ethyl
-C
2
H
5
propyl
-C
3
H
5
butyl
-C
4
H
9
Pentyl
-C
5
H
11
The branch is on carbon 3 and it has i carbon atom , the branch name is methyl
Name of the isomer is 3 methyl hexane
the
Example. h
Name the following isomer
Answer : 4-ethyl -1,5dimethyl octane
NB .
1. If they are more that one branches arrange them in alphabetical order.
2. If the are two or more similar branch use the following prefix
No of branch
Prefix
2
Di
3
Tri
4
Tetra
5
pent
Physical properties of Alkanes
Alkanes has the following physical properties
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1. They have low Melting and Boiling points . Smaller Alkanes are gases at room temperature while
other are liquid at room temperature . The Mpts and Bpts increase as the chain increase or size of
the molecule increases which increases IMF
2. They have low density compare to other organic compounds
3. They are insoluble in water and are soluble in other organic solvent like carbon tetrachloride
4. They have love viscosity compared to other organic compounds
Chemical Properties
Since Alkanes are saturated they are nonreactive but they react with substance that are reactive them self
E.g oxygen and radical
Combustion of Alkanes
They burn in air with clean, luminous, non sooty flame to give carbon dioxide and water. It absences of
enough air it undergo incomplete combustion. Flame is need to initiate the reaction
Complete combustion
It produces hot blue flame used for cooking
Alkanes + oxygen carbon dioxides + water
2C
2
H
6
+ 7O
2
4 CO
2
+ 6H
2
O
Incomplete Combustion
It gives yellow flame in candles and lamps
2C
2
H6 + 3O
2
4 C+ 6H
2
O
NB : A flame is needed to initiate the combustion reaction of Alkanes since it increases the kinetic and
speed of the molecules which make the bond to be broken during collision and start reaction
Substitution Reaction of Alkanes with Halogen
Ultra violet light rays are need to break the bonds in the halogens to initiate reaction. The reaction
produces a haloalkane
Substitution reaction with chlorine
Ultra violet light rays breaks the bond between the Cl-Cl to initiate the reaction. The Cl radical are very
reactive and substitute the hydrogen atom from the Alkanes and produce chloroalkane .
For examples.
CH
4
+Cl
2
CH
3
Cl + HCl the product is a mono- chloro-methane
The reaction can go further to produce dichloromethane
CH
3
Cl + Cl
2
CH
2
Cl
2
+ HCl
If chlorine is in excess the reaction can go further to produce trichloro-methane ( CHCl
3
) and later
tetrachloro-methane ( CCl
4
)
NB; trichloro-methane is mostly used in hospitals as anesthetics. All halogen can undergo similar
reaction as a chlorine.
Halo- alkanes has the same physical and chemical properties as Alkane burn they do not burn in air
Nitrification of Alkanes
Alkane reacts with concentrated nitric acid at high temperature of 300 degree Celsius to produce nitro
ethane and water
C
2
H
6
+HNO
3
C
2
H
5
NO
2
+ H
2
O
PREPARATION OF ALKANES
Fractional distillation of crude oil
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Crude oil as different liquids that are separated by fractional distillation. First gaseous alkanes are released
and later liquid and finishing with solids
CRACKING
Long chain are broken down into small chains and more useful one
TYPES OF CRACKING
Thermal crackling/pyrolysis
This is achieved by heating the long chains to the temperature of 140 degree Celsius and 700 degree
Celsius.
Long chain smaller chain Alkanes + Alkanes and hydrogen
Ethane used in polymerisation is produced in this way
Catalytic cracking
Long chain Alkanes are passed over a catalyst on the temperature of 450 to 500 degree Celsius
Long chains catalyst Smaller chain Alkanes + alkenes and hydrogen
Bio gas generator
Animal manure and straw are allowed to decay an aerobically to produce methane gas which can be used
for cooking
Laboratory preparation of methane gas
Methane is prepared by heating sodium ethanoate and soda lime
CH3 COONa (s) +NaOH (s) CH
4
+Na
2
CO
3
(s)
Uses of Alkanes
1. Used as fuel in motor vehicles eg gasoline
2. Used in cooking eg methane gas
3. Used in gas welding
4. Used for lighting purposes
5. Used as lubricants
6. Used as waxes
7. Halogenoalkanes are used as solvents in paint stripper
ALKENES
They belong to hydrocarbon. They contain hydrogen and hydrogen atoms only.
Alkene has C=C ( carbon -carbon double bond) as its functional group .
They are unsaturated hydrocarbon because of the presence of the double bond since other atoms can be
added across the double bond to become saturated.
Their general formula is C
N
H
2
N
Because of the presence of the double bond the smallest Alkene is ethene with two carbon atoms.
System of Naming Alkenes
It is the same as Alkenes but it only differs on suffix where it ends with -ene
No Of C
Formula
Structure formulae
Name
2
C
2
H
4
Ethene
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3
C
3
H
6
Propene
4
C
4
H
18
Butene
5
C
5
H
10
Pentene
6
C
6
H
12
Hexene
7
C7H14
Heptene
8
C8H16
Octene
9
C9H18
Nonene
10
C10H20
Decene
NB: according to IUPAC rules the position of the double bond should be included in the name
Start numbering from the near where there is a double bond
Example -C=C-C-C-C-C-C-C-C-C- dec-1-ene
C-C-C=C-C-C- hex-3-ene
C- C-C-C=C-C-C- hept-3-ene
Isomers of Alkanes
In isomer of Alkene the position of the double is included in the name
Examples
C=C-C-C-C Pent-1-ene
C-C=C-C-C pent-2-ene
2methyl but-2-ene
NB In molecule with functional group numbering should start to point nearest to functional group
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PHYSICAL PROPERTIES OF ALKENES
1. They have low melting and boiling points like Alkanes and smaller Alkenes are gases are room
temperatures
2. They have low density
3. They are insoluble in water
Compounds with two functional group of Alkenes are referred as alkadienes while these with three
functional group are referred as alkatrienes
The prefix di, tri are used to should the number of the functional group present
Example
C-C=C-C=C-C Hex -1,4 -diene
C=C-C-C=C-C=C-C Oct-1,4,6-triene
Cycloalkane
They are Alkene with ring and double bond
To name cyclo Alkanes the ring should be numbered so that the double bond is given number of 1 and
2 The direction of numbering is chosen so that the substitute receive the lowest number
Example
4 ethyl3methyl cyclopentene
NB The position of the double bond is not include because the position is already know to be 1 and 2
CHEMICAL PROPERTIES OF ALKENES
Combustion
Alkane burn with a sooty flame because they are unsaturated and contain more carbon atoms to
hydrogen ratio.
Complete combustion
Burn with hot blue flame forming carbon dioxide and water
Ethene + oxygen carbon dioxide and water
C
2
H
4
+ 3O
2
2 CO
2
+ 2H
2
O
Incomplete combustion
It burn with yellow cooler flame and produces carbon or carbon monoxides
C
3
H
6
+ 3O
2
3 CO + 3H
2
O
ADDITIONAL REACTION
Because of the presence of the double bond Alkene molecule makes them to undergo additional
reaction where on of the double bond breaks and other atoms can be added across the double bond.
Additional reaction with halogens
Bromine and chlorine instantly decolourising when added to Alkene. This shows that bromine and
chlorine are added across the double bonds.
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1,2 dichloroethane (C
2
H
4
Cl
2
)
With bromine
1,2 dibromoethane ((C
2
H
4
Br
2
)
Addition with potassium manganate (VII)
When added to Alkene, acidified potassium manganate(VII) change from purple to colourless .
Potassium manganate act as an oxidation agent adding oxygen across the double bond
Etha-1,2-diol
With Acidified potassium dichromate (VI)
It changes from orange to green. It acts as oxidising by adding oxygen across the double bond.
Addition of hydrogen chloride
The additional occur instantly across the double bonds
chloroethane
Concentrated sulphuric acid
When concentrated suphuric acid bulbed in alkene it is added instantly on the double bond
(Ethyl hydrogen sulphate (C
2
H
3
CH
2
SO
4
H)
Addition of Hydrogen (Hydrogenation)
The reaction takes place using a nickel catalyst under high temperature ranging from 140 to 400 degree
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Celsius to break strong bond in hydrogen molecule to form corresponding alkanes
Preparation of Alkene
They are prepared by cracking of Alkanes
Test for Alkenes
Chlorine water, bromine, acidified potassium dichromate (VII) , potassium chromate can be used to test
the presences of Alkenes because they is a change in colour during the reaction
Few drop of each is added to test tube contain samples. For example to test the presence of alkenes
using bromine solution, add 5 drops of bromine solution to 15 drops of samples or unknown liquids.
ALKONOLS
They belong to oxycarbons and they contain -OH ( hydroxyl group ) as their functional group The
General formula for the alkanols is C
N
H
2N+1
OH
System of Naming of alkanols the naming system is the same but the end name is -anol
No Of C
Formula
Structure formulae
Name
1
C
1
H
3
OH
Methanol
2
C
2
H
6
OH
Ethanol
3
C
3
H
5
OH
Propanol
4
C
4
H
9
OH
Butanol
5
C
5
H
11
OH
Pentanol
6
C
6
H
13
OH
Hexanol
7
C
7
H
15
OH
Heptanol
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8
C
8
H
17
OH
Octanol
9
C
9
H
19
OH
Nonanol
10
C
10
H
21
OH
Decanol
Classes of alcohols
They classified to the position of the functional group ( hydroxyl group) is attached in the molecule
Primary alcohols
The hydroxyl group is attached to carbon atom with two other hydrogen attached
Examples
Secondary alcohol
The hydroxyl group is attached to carbon attached to two carbon atom and one hydrogen atom
attached
Tertiary alcohol
The hydroxyl group is attached carbon no other hydrogen is attached, it is attached to carbon attached
to other two three carbons
NB: alkanols with two functional group are called alkadiol while with three function group are called
alkatriols
For example :
propan1,2 diol
Hexan-1,4,5 triol
ISOMERISM IN ALCOHOLS
Alkanols from propanol going down the group has more that one structure
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Example of isomer of butanol
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OH-C-C-C-C Butan-1-ol
Hexan-2-ol
methyl propan-2-ol
Physical Properties of Alkanols
Due to presence of OH which give to rise of hydrogen bonding and stronger intermolecular forces alkanols
has the following physical properties
1. Smaller alkanols are liquid at room temperature
2. They have high melting and boiling points
3. They are more viscous than Alkane and Alkene an these with more than one -OH are more viscous
4. Smaller alkanols are soluble in water but large alkanols from pentanol are insoluble while butanol is
partially soluble
Chemical Properties
Combustion
Alkanols burn completely in air because in their molecule they already oxygen
2C
2
H
5
OH + 3O
2
2CO
2
+ 3H
2
O
Reaction with sodium
The reaction produces hydrogen gas and sodium ethoxide
Ethanol + sodium sodium ethoxide + hydrogen gas
Reaction with ethanoic acid – Esterfication
The reaction takes place on the presence of concentrated sulphuric acid to initiate the reaction
Ethanol + ethanoic acid ethyl ethonoate (ester)+ water
C
2
H
5
OH + C
2
H
3
COOH C
2
H
3
COOC
2
H
5
+ H
2
O
REACTION WITH ACIDIFIED POTASIUM DICHROMATE (VI)
EXPERIMENT
Oxidation of ethanol to ethanoic acid by acidified potassium dichromate
Materials
Ethanol
Potassium dichromate (VI) 2 molar
Dilute sulphuric acid
A test tube in a rack
2 dropers
Syringes 5 ml
Test tube holder
Burner and matches
Thermometer
Procedure
a. Put about 5ml of potassium dichromate 2 m solution into a test tube
b. Add 2ml of dilute sulphuric acid
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c. Warm the mixture to about 50Cby putting the test tube onto burner
d. Add 1 ml ethanol to tube using a syringe
e. Leave the test tube standing for same time about 5 minute and observe any changes
Results
Potassium dichromate solution turns green from orange
Conclusion
This show that ethanol has been oxidised. Potassium dichromate acts as oxidising agent for ethanol. Ethanol
is oxidized in two stages
In the first stage ethanal is produced
C
2
H
5
OH acidified potassium di chromate C
2
H
3
CHO
Later is oxidize to ethanoic acid
C
2
H
3
CHO C
2
H
3
COOH
In general
Primary Alkanol Aldehyde as intermediate effect carboxylic acid
Reaction with Concentrated Sulphuric Acid – Dehydration
The reaction produces ethene since water is removed from the molecule. The reaction can be ether partial
dehydration to form an ether or complete dehydration to produce ethene
C
2
H
5
OH Ethene + H
2
O
NB: The relative size of –OH determine the flame colour, speed of reaction and solubility in water
The colour change from blue to yellow, the reaction slow down and the solubility decrease because as the
size of the molecule increase the relative size of OH decreases compared to the hydrocarbon chain
Uses of alkanols ( Ethanol)
1. They are used as motor vehicle fuel blending agent eg ethanol
2. Used in manufacturing of ethyl ethanoate and ester for fruit flavoring for sweets and drinks
3. As a solvent
4. In manufacturing of cosmetics
5. Is used as fuel for motor vehicles
6. As a social drink
Test for Alkanols
They are two tests for alkanols
Solubility test where water is used to test smaller chain alkanols and sodium test where a small piece of
sodium is plugged into a solution of Alkanols
PREPARATION OF ETHANOL
They are two ways how ethanol can be produced
By fermentation of sugar and dehydration of ethene from petroleum
FERMENTATION OF SUGAR
Fermentation is the process in which sugar and starch are decomposer anaerobically by bacteria and yeast in
process producing ethanol and energy. Bacteria produces a enzyme called sucrase which convert sucrose to
glucose /fructose .
Sucrose + bacteria glucose + fructose
Yeast produce the enzyme called Zymase which convert Fructose / Glucose to Ethanol and Carbon dioxide
Glucose Yeast Ethanol + Carbon dioxide + energy
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Ethanol is prepared my fermenting sugar with yeast.
The mixture of ethanol and maize husk is later undergo distillation process to produce ethanol
Hydration of Alkene
Ethene is mixed with a steam of water and is passed over a catalyst . Ethene is hydrated in the process
whereby a water molecule is added across the double bond to produce ethanol.
The product solution of water and ethanol is separated by fractional distillation
Why Production of Ethanol from Starch is Preferred that Hydration of Alkenes
The following are reasons
1. It uses renewable resources like corn and sugarcane while hydration uses oil which is non-renewable
resource.
2. It is good ways to use waste organic material like maize husks.
3. It uses cheap local available materials.
Disadvantage
i. Large volume is need to produce a litre of ethanol
ii. It is slow compare to hydration of ethene
iii. It does not produce pure ethanol compared to ethanol produced from hydration of ethene
CARBOXYLIC ACIDS
They are oxycarbons because they contain oxygen in their molecule.
Their functional group is carboxylate ( -COOH)
Their general formula is C
N
H
2N-1
COOH
In carboxylic acid number of a carbon is equal to N-1 because one of the carbon atom is attached to the
functional group.
For example
Write the molecular formula of carboxylic acid with N = 6
Since one carbon is in the functional group N = 6-1 = 5
The formula for that compound is C
5
H
2x5
COOH = C
5
H
11
COOH and the name is Hexanoic acid
System of naming
Like other homologous series the prefix is the same but the suffix is -anoic acid
No Of C
Formula
Structure formulae
Name
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0
H COOH
Methanoic acid
1
CH
3
COOH
Ethanoic acid
2
C
2
H
5
COOH
Propanoic acid
3
C
3
H
7
COOH
Butanoic acid
4
C
4
H
9
COOH
Pentanoic acid
5
C
5
H
11
COOH
Hexanoic acid
6
C
6
H
13
COOH
Heptanoic acid
7
C
7
H
15
COOH
Octanoic acid
8
C
8
H
17
OH
Nonanoic acid
9
C
9
H
19
COOH
Decanoic acid
Natural source of carboxylic acids
1. Citrus fruits example acetic acid
2. Milk
3. Vinegar
4. Anti bite
5. Tomatoes
6. Human sweat
Uses of carboxylic acid
1. In textile processing
2. Grain preservative formic acid
3. Additive to bread making calcium propanoate
4. Additive to butter butanoic acid
5. As food preservative eg benzoic acid
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Physical properties of carboxylic acids
Due to presences of Carboxylate as functional group which give the carboxylic acid strong intermolecular
forces and hydrogen bonding. The strong intermolecular forces and the hydrogen bonding tend to pull the
molecule more closer to each other carboxylic acid has the following are physical properties
1. The highly soluble in water
2. They are solid room temperature
3. They have high density
4. They have high viscosity
5. They have high melting point and boiling points
Note :
Small carboxylic acid exist as liquid while large one exist as solid at room temperature because they have
strong intermolecular forces and the hydrogen bonding tend to pull molecules closer to each other hence the
liquid and solid state while alkanols the intermolecular forces are pull electron closer top each other but no
like as in carboxylic acid hence most opf them exist as liquid at room temperature.
Alkane and alkenes they solely depend on weak intermolecular forces as which fails to pull molecules closer
to each other hence the gaseous state.
Carboxylic acid has the highest melting and boiling point because it has strong intermolecular forces called
hydrogen bonding that need a lot of energy to break them compared to alkanols, alkanes, alkenes.
Hydrogen bond is bond that is formed when hydrogen is bonded to most electronegative atom like oxygen,
fluorine , nitrogen. Hydrogen bond are stronger than intermolecular force but they are weaker that chemical
bonds.
In carboxylic acid oxygen pull electron toward itself as a result around oxygen tend to partially negative.
Carboxylic acid tend to form two hydrogen bonds while alkanols forms a single hydrogen bond
CHEMICAL PROPERTIES OF CARBOXYLIC ACIDS
Acid reaction
They are weak acids and they donate H+ from the functional group
C
2
H
5
COOH + NaOH (aq) C
2
H
5
COONa + H
2
O(l)
Ionization of water
Carboxylic acid ionises water and give a pH value less than 7 which is acidic condition
C
2
H
3
COOH + H2O C
2
H
3
COO
-
+ H
2
O + H+
CONDUCTIVITY OF CARBOXYLIC ACIDS
Because carboxylic acid ionise in water it conduct electricity compare to Alkanols, Alkene and Alkanes
which does not ionise in water.
React with alkanols (esterfication)
The reaction takes place on the presence of dilute sulphuric acid to initiate the reaction
Ethanol + ethanoic acid dilute sulphuric acid ethyl ethonoate (ester)+ water
catalyst
C
2
H
5
OH + C
2
H
3
COOH dilute sulphuric acid C
2
H
3
COOC
2
H
5
+ H
2
O
catalyst
NB: Esterfication is the process of preparing esters by reacting alcohol/ alkanols with carboxylic acids
How ethyl ethanoate is prepared
It is prepared by reacting ethanol with ethanoic acid in dilute sulphuric acid.
Using the distillation apparatus the mixture is distilled to produce ethylethanoate.
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The reaction is reversible and is at equilibrium the distillate contain excess ethanol and ethanoic acid to
remove excess acid sodium hydroxide is added to neutralise it and sodium ethanoate is produced in the
process.
Excess ethanol is removed by adding anhydrous calcium chloride to produce calcium ethnocide. . Ethyl
ethanoate is is removed by using separation funnel. Later the impure ethylethanoate is distilled at a
temperature of 74ºC to obtain pure ester
Uses of ester.
1. Artificial fruit flavouring
Name of ester
smell
Prepared from
Alcohol /alkanols
Carboxylic acids
Octyl ethanoate
Orange
Octanol
Acetic acid
Butyl ethanoate
Banana
Butanol
Ethanoic acid
Ethyl butanoate
Pineapple
Ethanol
Butanoic acid
Methyl anthranilate
Grape
Methanol
2 aminobenzoic acid
Isobutylpropanoate
Rum
Iso butyl alcohol
Propanoic acid
n-propyl acetate
Pears
n-propyl alcohol
Butanopic acid
Benzyl butyrate
Flower
Benzyl alcohol
Butanoic acid
2. Used as substitute for cotton
3. Used for bonding region in glass fibre
4. For making food roasting
5. For yatch nails
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6. For making perfume
Test for carboxylic acid
Acid test is used to test the presence of carboxylic acid
The following indicator can be used to test the presence of carboxylic
Indicator
Colour in acid
Colour in bases
Litmus
Red
Blue
Methylorange
Red
Yellow
Phenolphthalein
Colourless
Red
Bromothymolnblue
Yellow
Blue
FLOW DIAGRAM
Experiment
Identification of alkanes alkenes, alkanols and carboxylic acid
Materials
4 test tube in a rack
Ethanol labelled A
Paraffin labelled B
Ethanoic acid (vinegar) labelled C
Vegetable oil (cooking oil) labelled D
Sodium hydroxide or lime water (calcium hydroxide) m in a dropper bottle
Phenolphthalein solution in a dropper bottle
Distilled water
Procedure
a. Add 15 drop of distilled water in each test tubes
b. Add at least 10 drops of the test liquid A, B, C and D inn a separate test tube and shake
c. Record the results in appropriate inn the flow diagram
d. Rinse the test tube with water followed by distilled water
e. Select the soluble liquids only. Put 15 drops of sodium hydroxide /lime water into test tube followed
by two drops of phenolphthalein indicator. Add 6 drops of each soluble tested liquid in test
separately. Record the results in appropriate place in the flow diagram
f. Select those insoluble liquids only. Place 15 drops of potassium permangate or bromine solution inn
the two clean test tubes. Add 2 drops of each the liquids separately. record your results in appropriate
in the flow diagram
om
Identify the homologous series from each liquid A, B C and D
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In general the flow diagram look like this
POLYMARISATION
Monomers
They are small molecules that join together to form long chains
Polymer
They are long chain molecules made by joining together many small molecule called monomers
Co-polymer:
It is a polymer formed by the polymerisation of a mixture of two or more different monomers.
Types of polymers
They are two types of polymers
1. Natural polymers
2. Synthetic polymers
Natural polymers
They are biochemical molecules than have long change
Example of natural polymers
Polymer
Monomers
Where it is found
Proteins
Amino acid
Wool, silk, muscles , etc
Starch
Glucose
Potato, cereal crops
Cellulose
Glucose
Paper, wood dietary fibre
DNA
Nucleotides
Chromosomes , genes
Synthetic polymers
They are made by joining small molecule to form long chain molecules
Example of synthetic polymers
Polymer
Monomer
Where it is found
Poly ( ethene )
Ethene
Bags, washingup bowls , etc
Poly(chloroethne) PVC
Chloro ethene
Fabric coating, electrical insulation, etc
poly(phenylethene) ( polystyrene)
phenlethene
Toys, expanded polyestyrene
Polyesters
Ethene-1,2-diol and
benene1-2-dicaboxylic acid
Skirt, shirts , trousers
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Factors that Affect the Properties of Polymers
It depends on the type of the molecular chain it contain
1. Chain length: they have long chain with an average of 50 monomer units. . The chain length determines
the melting and boiling points of polymers.. Most polymers have high melting and boiling point
compared
2. Intermolecular forces: The size and strength depend on the chain type and molecule type that made the
monomer. Polymers formed from condensation polymerisation tend to have high melting and boiling
point compared to polymers that come from additional polymerisation since they tend to form hydrogen
bond
3. Branching : non branched polymers are closely packed while branched polymers are not closely
packed. Branched polymers melt easily and have weak tensile strength.
4. Cross- linking: they are rigid and form hard material than straight chain polymers.
Types of polymerization
They are two type of polymerization
Additional polymerisation and condensation polymerisation
Additional Polymerisation.
One type of monomer molecules undergoes addition reaction repeatedly.
Molecules with double bonds are particularly used in additional polymerisation E.g ethene . One the double
bond break enabling the free bond to link with one another forming a chain.
Examples of additional polymers
polyethene polystyrene PVC poly(propene) Polytetrafloroethane (PTFE) Acrylic
POLYETHENE
It is made using ethene as a monomer. The ethene molecules undergo reaction where the double bond break
and join to another ethene molecule to form polythene .
Polythene molecule
Properties of polyethene
1. They are easily moulded
2. It is excellent electrical insulator
3. It does not corrode
4. It is tough
5. It is not affected by weather
6. It is durable
Types of polythene
Low density polythene
They are boiled at 105 degrees Celsius and soften in boiling water.
It has branched chain which made them not to be closely packed.
They are fairly readily melted and easily b deformed.
They always soften in hot water
Used.
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1. For making films, sheeting for bags, wrappers ,
2. For making moulded article like washing –up bowls and squeezy bottles
High density polythene
1. They are made at low temperature and pressure under the influence of catalyst.
2. They are closely packed because they have few branches
3. They do not melt easily and they are rigid.
Uses.
They are used for moulding bleach bottles, milk -bottle crates
Poly(propene) (Polypropylene)
The use propene as a monomer. The double bond break which give a way for other bond to be formed by
joining the propene monger to form poly (propene).
Poly propene
Properties of poly propene
1. Its chain are packed together closely
Uses of poly propene
1. They are used in mouldings and film
2. they also used to made fibre
3. they are used to make ropes
Poly( chloroethene) ( Polyvinyl)
Chloromethane undergo addition reaction to produce PVC
poly vinyl chlorine
Characteristics of PVC
1. They are versatile
2. They have strong IMF due to presence polar c-cl bond. And this made them very strong material
uses:
1. Used in coating fabric and covering wires and cables.
2. used in making water pipes
Other additional polymers
Polymer
Monomer
Monomer
formula
Monomer
common
Properties
Uses
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name
poly(phenyle
thene
phenylethene
C6H5CH=CH2
Styrene
Brittle (but
cheap)
Expanded [polystyrene for
insulation
plastic toys
poly(tetraflu
oroethene)PT
FE
tetrafluoroeth
ene
CF
2
=CF
2
Tetrafluor
oethylene
Very stable
anti-stick
properties
Non-stick coating on pans
insulator
poly(methyl-
2-methyl-
propnoate)
(perspex)
Acrylic
methyl-2-
methyl-
propnoate
CH
3
C-COOCH
3
CH
2
Mythyl
methacryl
ate
Transparent
As substitute for glass
Poly(
propenenitril
e) Acrilan
propenenitrile
CH
2
=CH
C
N
Acylonitr
ine
Strong
fibre
properties
Making textile ( wool)
Condensation polymerisation
It involves reaction of two monomer where a small molecule mainly water molecule is released to form a
bond between the two large molecules.
It usually happens on a two types of monomer.
Examples of condensation polymerisation
Nylon
Two monomer 1,6 diamono hexane and hexa-1-dioyl chloride join each other by eliminating hydrogen
chloride by amide link.
Hexane-1 , 6 diamine hexa-1-6 dioyl chloride
The two monomers join each other as follows
In absence of the hexa -1,6 dioylchloride as monomer, Hexanedioic acid can be used can be used and the
reaction produces water like
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TERYLENE
Terylene is made by condensation polymerisation of monomers bezene 1,4 dicarboxylic acid
[HOOC(C
6
H
4
)COOH] and ethano1,2 diol [HO(CH
2
)
2
OH] by an ester link
Benzene 1,4 dicarboxylic acid Ethan-1, 2 diol
The function group of is the one that take place in polymerisation
Other examples of condensation polymerisation
bakelite - terylene - urea-formaldehyde
Many substances which occur in living organisms such as proteins and polysaccharides polymers
Plastics ( Synthetic polymers )
Plastics are a material that comes mobile when heated.
Types of plastics
Thermoplastic ( Thermo softening plastic ):
They are made by straight chain polymers.
The chain can slide over each other
Properties of thermosetting plastics
1. It is a material which softens on heating and hardens on cooling , the process being capable of
repetition
2. They are flexible and do not easily break
3. Melt at low temperatures
4. Can be folded into a shape after they are made
Examples are:
Lucite , Teflon, nylon, polystyrene , plastic eye glass frames
3 Thermosetting materials:
They are made by strong bond between the chain to form cross-linking structure
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Properties
1. They will break or char down at high temperatures rather that melt
2. They are rigid and break rather than bend
3. Once they are moulded into a shape while they are made they can no be changed later.
NB thermosetting plastics materials can acquire plasticity during the first heating and cannot be soften on
re-heating . This is due to a formation of a molecular network structure between the long chain molecules
which produce a rigid 3- dimensional network.
Examples are:
Bakelite- radio cabinets –Formica –polyesters –cassein.
Properties of plastics
1. Some are hard, light and durable and can be used in the manufacturing of the body of a car.
2. Some have excellent heat and electrical resistance and can be used as insulators.
3. Some are non- biodegradable.
4. They are water-proof
5. Some are flexible
6. Some soften upon being heated.
Uses of plastics
1. Some are used in textile industry e.g. nylon and polyesters
2. As plastic bags, containers, water piping, film and sheets such as pol(ethylene)
3. For carpets, bearings such as nylon
4. For tyre cord soft drink bottles such as terylene
5. For buttons, knife handles, pens, switches, camera & telephone equipment such as Bakelite
Laboratory preparation of plastic nylon ( polamide )
Procedure:
a. Dissolve about 1 ml of decan-1,10-dioyl chloride in 10ml of tetrachloromethane in a 100ml beaker.
Add a trace of azobenzene to colour this solution.
b. In a separate beaker , make a solution of 1g of 1,6-diaminohexane in 20ml of sodium hydroxide
solution.
c. Carefully pour the solution from (b) on top of the solution made in (a ) , taking care to avoid
mixing the two solutions.
Result: nylon forms at the interface of the two solutions.
Advantages of synthetic polymers over metallic materials
1. Some plastic are hard light and durable and can be used to make body of cars while some metals are
hard , heavy and not durable
2. They have excellent heat and electrical resistance compared to metals as a results they are used and
insulator while metal are good conductor of heat and electricity and can not be used a insulators
3. They are water proof and can not attacked by rusting while metals can be easily attacked by rusting
like iron
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4. They melt at low temperature hence they can be easily recycled while metals melt at high
temperature hence expensive to recycle
5. The are cheaply produced while metals are expensive to produce
Disadvantages of synthetic polymers
1. Some are toxic and can cause death to animals and even people
2. They pollute environment since most of them are un bio degradable nor photo degradable
3. Some can not be recycled
4. They release chlorofluorocarbons during manufacturing that damage ozone layer
5. They are use non renewable resources like petroleum once depleted they will no longer exist.
Advantages of natural polymers
1. All can be recycled
2. They com from renewable resource like trees
3. The are environmental friendly
4. They are non toxic
5. They are biodegradable and photo degradable
Disposal of plastics and wastes.
Burning them.
Plastics are burnt in incinerator. But it has the disadvantage that it increase pollution of air through releasing
of chlorofluorocarbon into the atmosphere
Using landfills
Plastic are buried in the soil. The disadvantage ids that plastic take long time to decompose and they
pollute the soil. Any pant that grow on that landfill of plastic dries prematurely
Re-cycling
Plastics are recollected and heated in the factory and made into new one.
The disadvantage is that other plastic can not be recycled and the process is expensive
Exercise
2003
Question 5
a. (i)Name the compound C
7
H
15
OH
ii. What is the general formula of the compound in 5 a.i ?
iii. Draw the structure of the C
7
H
15
OH
b. Figure 4 shows one indigenous way of preparing alcohol
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i. Name the process illustrated in figure 4
ii. Name the alcohol collected in receiving bottle
iii. The alcohol in figure 4 is produced by fermentation
1. Define fermentation
2. Write the word equation from fermentation
c. Ethanoic acid (CH
3
COOH) react with ethanol (C
2
H
5
OH)according to the following equation
CH
3
COOH + C
2
H
5
OH conc. Sulphuric acid CH
3
COO C
2
H
5 +
H
2
O
Heat
i. What is the name of this reaction
ii. Name the two products of this reaction
iii. Give one use of CH
3
COO C
2
H
5
d. Draw and name the isomers of pentane (C
5
H
11
)
2004
a. State one use of each of the following polymers
1. Plastics
2. Carbohydrates
b. State any two ways of disposing plastics s
c. The following are structural formulae of four molecules with molecular formula C
4
H
8
i. Name the molecules 1 and 2
ii. Which two structures is conformation of each other?
d. Table 4 shows the molecular formulae and the boiling point of some compound
TABLE 4
Compound
Molecular formula
Boiling point (ºC)
A
C
2
H
4
-104
B
C
2
H
5
OH
79
C
CH
3
COOH
118
D
H
2
O
100
E
C
2
H6
-89
i. Which compounds in the table are hydrocarbons
ii. Which compounds in the table are soluble in water
iii. Which compounds in the table are gases at room temperature
iv. Explain why the boiling point of compound D is higher than the boiling point of E
v. Describe a test which can be done to distinguish C and D
2005
Question 5
a. Figure 5 is a diagram showing how ethanol and ethanoic acid are prepared
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i. Give the names of
1. Gas A
2. Process B
3. Reaction C
ii. Name the substance that is used in reaction C
b. The following are structures of some organic compounds
i. Name compound A
ii. Which compound is soluble in water? Give a reason for your answer
iii. Write letter reperesneting any three compounds that would not react with potassium , group 1
element
iv. Which of the two compound A and E would have lower boiling point? Give a reason for you
answer?
v. Compound C is monomer. Write the equation to show it polymerisation
c. Describe a test that could be used to distinguish compound D from E
d. Write the isomers of the substance A
e. Give two advantages of thermoplastics
2005
Question 1
a. Given below are general formulae of some homologous series reperesented by letters P, Q, R and S
P: C
N
H
2
N
Q: C
N
H
2N+2
R: C
N
H
2N+1
OH
S: C
N
H
2N+1
COOH
i. Name the homologous series represented by letter Q and S
ii. Which general formulae represent hydrocarbons
iii. Draw the structure of compound with three carbon atoms in homologous P
iv. Name the compound drawn in 1 a. iii.
v. Explain how a compound of homologous series Q could be distinguished from a
compound of homologous series R
b. i. Write down all structural isomers of pentane
ii. name the isomers
c. Ethene (c2h4) react with bromine in additional reaction.
i. Draw the structure of the product formed
ii. Name the product formed
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iii. Why is additional reaction important in industries? Give two reasons
2007
Question 1
a. Polymerisation of ethene can be summarised by the following equation
i. Name the polymerisation represented by the equation
ii. Describe how the polymer is formed from ethene molecule
iii. Give two examples of artificial polymers
b. The following are chemical formulae of some organic compound :
A. CH
3
CH
2
CH
2
CH
2
OH
B. CH
3
CH
2
CH
2
CH
3
C. CH
3
CH
2
CH
2
COOH
D. CH
3
CH
2
CH=CH
2
E. CH
3
CH
2
H
2
CH
3
i. Identify one compound which is an alkanols
ii. Which compound belong to the same homologous series
iii. Explain why solution of compound C conduct electricity
iv. Draw a full structure of compound D
v. Name compound D
c. i. Give three differences between thermosetting and thermoplastic polymers
ii. State two ways of disposing plastics waste to avoid pollution
d. Give three advantages of plastic over metallic materials
Question 7
a. Draw full structure of ethanol (C
2
H
5
OH) and water (H
2
O)
b. Explain the difference in boiling between water and ethanol
c. With an aid of diagram explain, describe an experiment tha can be done to separate a mixture of
ethanol and water
Question 3 PPII
Construct a flow diagram that could be used to identify acetic acid, ethanol, hexane and hexane using
test that make use of distilled water, bromine solution, sodium hydroxide solution and phenolphthalein
solution.
2008 a
Question 5
a. Define isomer
b. i. Draw the structural formula for two isomers of butane (C
4
H
10
)
ii. Name the isomers draw in 5bi
c. Figure 5 is a diagram of experiment set up.
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i. Name the process that could occur in the flask
ii. Write down the balance equation of the process named in 5.c.i
d. State any two disadvantages of synthetic polymers
e. Table 2 shows boiling point and solubility of some alkanols
Name
Boiling point
Solubility
Methanol
65
Most soluble
Ethanol
78
Soluble
Propanol
97
Partially soluble
Butane
117
Insoluble
i. Explain why boiling point of alkanols increases from methanol to butanol
ii. Explain why ethanol is more soluble in water than propanol
f. i. Complete the following equation to show reaction between methane and chlorine
CH
4
+ Cl ____________________
ii. Name this type of reaction
iii. give one use of alkanes
2009
Question 5
a. State any three use of ethanoic acid
b. i. Why ethanoic acid is regarded a electrolyte?
ii. Write the ionisation reaction of ethanoic acid (CH
3
COOH) in water (H
2
O
iii. Why does sodium metal react with ethanol in the same way as it does with water?
c. i. Write the general formulae for carboxylic acid
ii. What is the formula of smallest carboxylic acid?
iii. How would the boiling point of butane compared with smallest carboxylic acid
iv. Explain your answer to 5 ciii.
2010
Question 4
a. i. What are polymers
ii. mention any two uses of polyethene
b. Give any three properties of plastics
c. i. Give three properties of carboxylic acids
ii. Mention any natural sources of carboxylic acid
d. i. Define isomers
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ii. Draw the structural formulae of butanol (C
4
H
9
OH)
iii. write the condensed formula of pentane
e. Ethanol(C
2
H
5
OH) reacts with ethanoic acid(CH
3
COOH) according to the following chemical
equation
C
2
H
5
OH + CH
3
COOH conc. Sulphuric acid __________________ + H
2
O
i. Compete the equation
ii. Name the process in which ethanol reacts with ethanoic acid
2011
Question
a. i. Give any two properties of polymers
ii. Explain how condensation polymerisation is formed
b. Mention any two properties of alkanols
c. Ethanol (C
2
H
5
OH) changes to ethanoic acid (CH
3
COOH) in the presences of atmospheric oxygen
i. What is the function of atmospheric oxygen
ii. Write the equation for the reaction
d. Figure 3 shows the structure of some organic compound P, Q R and S
I. Name the organic compound labelled P and Q
II. Identify the structure which is isomers to compound Q
III. Give a reason for you answer in 5 c ii.
IV. Explain how the compound p could be distinguished from compound r using bromine solution
2012
Question three
The following are the general formula of organic compound A and B
A: C
N
H
2N+1
COOH
B: C
N
H
2N+1
OH
a. To which family does compound B belong
b. Mention any three properties of compound A
c. State any three uses of compound B
d. Mention the product formed when compound A and B react
e. Work out the molecular formula of the compound A if n is 5
f. Describe how compound A could be distinguished from compound B
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OSCILLATION AND WAVES
Oscillation
It is to and fro movement of the body on a fixed point
Example of oscillating object
Pendulum
It has swinging mass that move to and fro on a fixed
Vibrating pendulum
Cantilever
It made of ruler clumped to a table as below
The spring
Characteristic of oscillation system
Vibrating object has the following characteristics
1. Amplitude ( x )
It is the maximum displacement of a vibrating particle from its rest position. The unit is cm or m
2. Period ( T )
It is the time taken for 1 complete vibration or cycle. Its unit is second
T= total time taken/ no. of complete vibration.
3. Frequency ( f )
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It is the number of complete vibrations or cycles performed in a second.
Its unit = cycles /second or vibrations / second.
The special unit is called Hertz ( Hz )
1 Hz = 1 cycle / sec.
How to measure the frequency of an oscillating object
The following steps are followed
1. Set the pendulum for example or vibrating string. Pull it aside while your time has set to zero of
stopping clock. Release the pendulum about 10cm.
2. As the oscillation swing bob reaches you or starting position count three two and one then zero
3. Once you count zero start the watch at the same time and continue counting until you reach 10.
4. Repeat with another set of reading almost three time and find the average of the readings.
Damped Oscillations:
These are dying oscillations.
Due to friction the oscillations do not continue forever as a result it dies gradually
Friction is present in two forms here:
1. Rubbing between the holding clamp and the moving spring ,string or ruler.
2. Air resistance to the moving part in each system.
The overall effect is that the swings die away as the total mechanical energy ( P.E + K.E ) , gets less and
less. The amplitude decreases with time.
Factors affecting Frequency of Oscillating Systems
Spring
1. Mass on spring
Experiment 1. Investigating effect of mass on the frequency of oscillating pendulum
Materials
Spring
Masse like 50g, 100g 250g and 250g
Clamp stand
Ruler
Stop watch
Procedure
a. Hang the spring on the retort stand and hang a 50g mass
b. Pull the spring down ward about 10cm and release it
c. Record the time taken to complete 10 completed oscillations in the table below
d. Repeat procedures a to c with 100g, 150g, 200g, and 250g
Table of results
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Mass on the spring(g)
Time taken for 10 complete
oscillation
Number of vibration per second
(frequency) number of vibration /time
50g
100g
150g
200g
250g
e. Plot the graph of mass against frequency
Observation of results.
Small mass has take a short time to complete one circle since small effort is applied compared to heavy
mass which need a lot of force to be pulled which increase time hence reducing the frequency
2. Material of spring :
some material are elastic and are pulled widely which means that it take lengthy time to come back to
original position while other are elastic which mean they take short period to complete its circle
3. Number of springs or the strength of the spring:
Increasing the number of springs increases the force that pulls the mass up and down compared to
small number of springs.
Factors that do not affect the frequency of the spring
Length of spring
Changes in amplitude
Factors that affect the frequency of the pendulum
Length of pendulum
Experiment2. To investigate the effects of changing the length of pendulum on frequency of vibrating
pendulum
Materials
A clamp
60cm string
1 mitre ruler
A mass of 50 g
A stop watch
Procedures
1. Set the apparatus as shown below
2. Tie the string to a clamp so that the distance should be 40cm from the clamp hand to mass
3. Pull the mass to one side about a distance of 10 cm and leave to vibrate freely
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4. Record the time taken to make 10 complete vibration
5. Calculate the number of vibration performed in 1 second
6. Repeat procedure 2 to 5 with 30, 20 and 10 cm length of string and record your answer in the
table below
Length of string (cm)
Time take for 10 complete
vibration
Number of vibration per
second (frequency)
7. Plot the graph of vibration per second /frequency against length of a string
8. Plot the graph of 1/frequency against length
Observation / conclusion
The frequency of pendulum decrease as the length increases because the bob take long period to complete on
oscillation and also the effect of air resistance is high which reduces the frequency of oscillating pendulum
Material of pendulum: It the string is E.g wire
Experiment effect of change in amplitude on frequency of oscillating body
Materials
A string
Meter ruler
Stop watch
Mass 50g,
Clamp stand
Stop watch
Procedures
1. Set the apparatus as shown below
2. Tie the string to a clamp so that the distance should be 50cm from the clamp hand to 50g mass
3. Pull the mass to one side about a distance of 10 cm and leave to vibrate freely
4. Record the time taken to make 10 complete vibration
5. Calculate the number of vibration performed in 1 second
6. Repeat procedure 2 to 5 with 15cm, 20, 25 and 30cm amplitude and record your answer in the
table below
Amplitude (cm)
Time take for 10 complete
vibration
Number of vibration per
second (frequency)
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7. Plot the graph of vibration per second /frequency against amplitude
Observation / conclusion
As the amplitude increases the frequency decreases because the bob takes long period to complete
oscillation.
Factors that do not affect frequency
Experiment 3
Effect of mass of the frequency of oscillating pendulum
Materials
A string
Meter ruler
Stop watch
Masses 50g, 100g, 150g, 200g 250g
Clamp stand
Stop watch
Procedures
1. Set the apparatus as shown below
2. Tie the string to a clamp so that the distance should be 50cm from the clamp hand to 50g mass
3. Pull the mass to one side about a distance of 10 cm and leave to vibrate freely
4. Record the time taken to make 10 complete vibration
5. Calculate the number of vibration performed in 1 second
6. Repeat procedure 2 to 5 with ,100g 150g, 200g, 250g and record your answer in the table below
Mass of the bob (g)
Time take for 10 complete
vibration
Number of vibration per
second (frequency)
1. Plot the graph of vibration per second /frequency against length of a string
Observation / conclusion.
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A s the mass increase the frequency remain the same. This show that mass does not affect the frequency of
oscillating bodies
Small change in amplitude
Cantilever
Factor affects it
I. Mass on end: The frequency increase with decrease in mass and decrease with decrease in mass.
II. Material of cantilever: Some material are rid and die easily compared to flexible materials
III. Length of the cantilever: The frequency increase as the length decreases and decreases the length
increases.
Factor that do not affect the amplitude of cantilever
Amplitude changes
WAVES
A wave is any disturbance that carries energy and momentum from its source.
Note: The particles of the medium do not move, but they just oscillate ( vibrate ) , passing on energy to next
particles.
Types of Waves:
There are two types of waves
Mechanic wave and electro magnetic waves
MECHANICAL WAVES
This is waves that require medium for their propagation
For mechanical wave to move it requires the following things:
1. An elastic medium that can be disturbed
2. An energy source to provide a disturbance to the medium
3. Some physical connection through adjacent particle of medium that can influence each other
TYPE OF MECHANICAL WAVES
There are two type of mechanical wave
1. Longitudinal Waves
These are waves in which the direction of vibration of particles is parallel to direction of a wave.
It is formed when the particles of the medium are displaced in the direction parallel to the direction of the
travel wave.
Example mechanical wave
1. Sound wave.
2. Vibrating spring
How longitudinal waves are formed
They are formed when an object vibrates; air particles near it vibrate too creating a series of compressions
and rarefactions.
NB. Compressions are high-pressure regions.
Rarefactions are low-pressure regions.
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2. Transverse Waves:
These are waves in which the direction of particles is perpendicular ( at right angles ) to the direction of a
wave.
Examples transverse wave
Water waves
Light waves.
It happens when periodically repeating the disturbance of this type transmitted through the medium
CHARACTERISTICS OF WAVES:
1. A wave transfer energy but there is no net transitional of matter
2. Mechanical wave have particles with inertia and elasticity
3. The particles oscillate in the same way as the neighbours do
Difference between longitudinal wave and transverse wave
Transverse wave
Longitudinal wave
Oscillation are at right angles to the direction of
wave
Oscillation are in line with direction of wave
Can travel through a vacuum
Can not travel in vacuum since it need medium
Produces crest and trough
Produces compression and rarefaction
TERMINOLOGIES USED IN WAVES
Wavelength ( λ )
It is the distance between two successive crests or troughs or is the distance between the particle that are in
phase. The unit is cm or m .
The symbol is λ ( lamda. )
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Wave Speed ( V)
It is the distance covered by a wave per unit time ( second )
The unit can be m / s or km /s.
Phase ( E)
Is the position and motion of particle at instant having the same disturbance
Particles are said to be in phase if they have the same displacement , speed, and move in the same direction.
Things are said to out of phase ( in anti-phase ) if they have different displacements , speed or move in
different directions .
Frequencies
Is the number of complete wave generated per second
Amplitude (a)
It is the maximum displacement of a particle of the particle from their equilibrium position
Period (T)
It is the time require for one wavelength to pass point. It is the time for the required for a disturbed particle
by a wave to execute one complete oscillation
Relationship Between Frequency ( f ) and Periodic Time ( T )
f=1/T: T = 1/f
Proof:
Suppose a vibrating object ( pendulum ) makes 5 vibrations in 4 seconds.
- This means f = 5 vibrations / 4 seconds = 1.25Hz
-This also means T = 4 seconds / 5 vibrations = 0.8 second.
f = 1/ T = 1 / 0.8 = 10/ 8 = 1.25 Hz
T = 1/ f = 1/ 1.25 = 100 / 125 = 0.8 second.
WAVE EQUATION
Wave speed ( V ) = Distance / Time = Wavelength / Periodic Time
V = λ / T
V = λ / 1/f = λ f
V = λ f
V = λ f is called Wave Equation. It shows the relationship between V , λ and f.
λ = V / f; f = V / λ .
Examples.
1. A radio wave has a frequency of 12MHz and a wavelength of 25m . Calculate the velocity of the wave.
2. The speed of a radio wave is 300 000km/ s. If the radio broadcasts on 90.2 MHz, what is the
wavelength of the radio wave?
Solutions:
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a) Convert all the units into standard units i.e in Hz , m , m / s .
V = λ f = 25 x 12MHZ = 25 x 12 000 000Hz = 300 000 000 m / s
2 V = λ f Then λ = V / f
λ = V / f
= 300 000 km / s
90.2 MHz
= 300 000 000 m/ s
90 200 000Hz
= 3.3 m
Electromagnetic Waves
These consist of a disturbance in the form of varying electric and magnetic fields. The electric and magnetic
fields oscillate at right angles to the direction and at right angles to each other.
No medium is necessary and they travel more easily in a vacuum than in matter.
Examples are : light ( VIBGYOR ) –Radio signals (or waves)
- X-Rays – Gamma rays
All electromagnetic waves have the following properties:
1. Can travel through a vacuum.
2. Are transverse.
3. Have a velocity of 3 x 10
8
m / s
4. Can be refracted\
5. Normally travel in straight lines.
6. Can be reflected.
7. Undergo diffraction
8. Carry no charge
9. Obey the wave equation.
10. Undergo interference
11. Transfer energy from one place to another
Some Properties Of Mechanical Waves:
Diffraction:
It is the bending or spreading out of waves as they pass around obstacles or through an opening ( gap ) into
two regions
The greatest diffraction happens when the gap is narrow and the and the wavelength is large enough
Water waves passing through the gap a spread out in all directions and the wave fronts produced are circular.
Note: Diffraction is only significant if the size of the gap is comparable ( about the same size as ) with the
wavelength of the waves.
If the gap is wider compared with the wavelength , waves continue straight on , some spreading out occurs
but it is less obvious .
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Diagram showing less diffraction when the gap is wide.
Note: When designing harbor, engineers apply the knowledge of diffraction by making a model.
Diffraction in double slit
When the wave passes though the double slit it diffract and later interfere on another forming a big wave.
The big wave are formed because interference is constructive since the waves have the same amplitude,
same frequency ate constant phase as a result they reinforce each other as the effect the amplitude double
Interference:
It is the combination of waves to give a larger or small wave.
It is formed when two or more wave from different sources re superposed.
At points where a crest from one wave arrives at the same time as a crest from the other wave , a bigger
wave is formed. The waves are said to be in phase.
At points where a crest and a trough arrive together , they cancel out ( if their amplitudes are equal ) .The
waves are exactly out of phase due to traveling different distances.
TYPES OF INTERFERENCE:
Constructive Interference:
It happens when two waves superposed with the same amplitude, the same frequency at constant phases.
Waves reinforce one another the effect the amplitude doubles
A crest of one wave coincides with the crest of another wave.
Diagrams showing constructive interference.
Destructive Interference:
It is formed by superposing of two waves that are out phase but of same amplitude. Waves cancel one
another . A crest of one wave coincides with a trough of another wave.
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Refraction:
It is changing of direction of the wave as it is passing from one medium to another. The bending occur on
the boundary of the two media
Reflection:
It is the bouncing back of a wave
The wave that approaches the boundary is called the incident wave and the one that bounces back is the
reflect wave.
LIGHT
Definition of term
1. Principal axis(PA): I is imaginary line that passes through the centre of a lens and is perpendicular to
the lens
Optical centre ( c ) : It is the point in the lens where the principal axis and the plain of the lens meet
Focal length f: Is the distance between the principal focus to the optical point
Principal focus: F Is the point through which ray parallel to the principal axis converge with the convex
lens
Image distance v : It is distance between the image and the lens
Object distance u: Is the distance between the lens and the object
Drawing construction rays
It helps to determine where image will be formed by the convex lens.
How to construct ray diagram
1. Draw the lens and the principal axis perpendicular to each other like this
2. Draw the ray from the top of the object travels through the centre of the lens
3. Aray from the top of object parallel to principal axis to strike the lens. On emerging from the lens it
should pass through principal focus F on the other side of the lens
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4. Draw the ray from the top of the object passing through F and should travel parallel after emerging from
Len
CHARACTERISTICS OF IMAGES FORMED BY A CONVEX LENS:
1. Object between lens and F:
The image is
1. Behind the object.
2. Virtual.
3. Erect.
4. Magnified.
2. Object at F:
The image is at infinity.
3. Object between F and 2F:
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The image is;
i). Beyond 2F,
ii). Real
iii). Inverted
iv). Magnified.
4.Object at 2F:
The image is,
i. At 2F
ii. Real
iii. inverted
iv. Same size as object.
5. Object beyond 2F:
The image is
i) Between F and 2F
ii) Real
iii) Inverted
iv) Diminished
DIFFERENCE BETWEEN THE REAL IMAGE AND VURTUAL IMAGE
Real images
Virtual image
Incase of the mirror, the image lies behind the
mirror
Incase of mirror, the image lies in front of the
reflecting surface
The light rays at a focal point behind the mirror
The light ray meets at a focal point in front of
the mirror
It appears inverted ( depending on the type of
lens e.g convex , concave )
It is usually appears erect
It can be obtained on a screen
In can’t be formed on a screen
In case of lens the image lies on the other side
of the lens
Incase of lens the image lies on the same side
of object
THE LENS FORMULA
1/u +1/ v = 1 / f
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Example:
An object 2cm high stands on the principal axis at a distance of 12 cm from a convex lens. If the focal
length of the lens is 9 cm, what is the position and nature of the image?
Solution:
1/u +1/ v = 1 / f: But u =12 cm; f = 9 cm; v = ?
1 / v = 1 / v – 1 / u
= 1 / 9 – 1 / 12
= 4 - 3 / 36
v = 36cm
Calculating Image positions Using Lens Formula:
In using the equation, note the following:
1. If the lens is convex , f is taken as positive.
2. If the lens is concave , f is taken as negative.
3. When v is positive , the image is real.
4. When v is negative , the image is virtual.
Note:
The equation can also be used for curved mirrors.
Converging lenses and mirrors have positive focal lengths.
Diverging lenses and mirrors have negative focal lengths.
Example:
An object 2cm high stands on the principal axis at a distance of 9cm from a convex lens. If the focal length
of the lens is 6cm, what is the position and nature of the image?
Solution:
1/u +1/ v = 1 / f: But u =9cm; f = 6cm; v = ?
1 / v = 1 / v – 1 / u
= 1 / 6 – 1 / 9
= 1 / 18
v = 18cm
the image is rel and inverted.
Measuring the Focal Length of a Convex Lens
i . Plane mirror Method
1. A plane mirror is placed behind the lens and a bright object in front.
2. The object is in this case a set of illuminated cross-wires, and its position is adjusted until its sharp image
appears on the screen beside the object.
When this happens , rays of light from the object are parallel when they strike the mirror and are reflected as
a parallel beam.
Light from the object must travel back along nearly the same path and hit the mirror normally.
The object is then at the lens principal focus ( F ).E.g The cross-wires and the card mark the position of the
principal focus of the lens.
Distant Object Method:
1. A window across a large room is used as an object.
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2. U is large and the rays enter the lens almost parallel to each other.
3. Move the lens until a sharp image of the window is formed at the other side of the room or screen.
4. The distance between the lens and the screen is roughly f .
5. Measure the length of the lens – screen distance using a ruler
HOW TO FIND THE FOCAL LENGTH OF CONVEX LENS USING GRAPHICAL METHOD
MATERIALS
Convex lens
Lens holder
Screen
Candle
Matches
Meter ruler
Procedure
a. Arrange the apparatus as shown below
b. Light the candle
c. Move the object at a distance of 15cm away from the lens
d. Move the screen until you focus the image on the screen
e. Measure the image distance from the screen to the lens and record on the table below
f. Repeat procedure c toe with 20cm , 25cm, 30cm, 35cm and 40 cm length
g. Record the results in the table below
Object distance
(U)(cm)
Image distance ( V)
(cm)
V+ U
1/V
1/U
15
20
15
20
25
30
35
40
Measurement of Object and Image Distances:
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The lens is set up in front of an illuminated object so that a real image is formed on a white screen placed on
the opposite side as shown below
1. Move the object to different distance from the lens o vary the value of u so that the screen is also
adjusted to give a new focused image on the screen.
2. Measure the distance of u and v using a ruler
Optical Instruments
These are instruments which use light.
The Projector
The projector consists of concave mirror, illuminate source of light , condenser lens , slide or film lens and
screen ( wall or white surface ).
Optical ray diagrams
The object is placed inverted to produce upright real image
The slide projector
Parts of projector and their function
The Projection Lens
It is mounted in a sliding tube , so that it may be moved to and fro.
It is used to produce a focused, magnified real inverted image on the screen.
It has a long focal length.
Concave Mirror
It reflects light back.
Condenser
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It consists of two plane-convex lenses of shorter focal length .
It collects light which would otherwise spread out and be wasted, and
It causes light to converge through the slide on to the projector lens in order to obtain a brilliant picture on
the screen.
The condenser converges and condenses ( concentrates ) much light onto the slide or film
Lamp/ bulb
The bulb or lamp is either a carbon electric arc or a quartz iodine lamp to give a small , but very high
intensity source of light.
It is situated at the centre of the curvature of a small concave mirror.
It is used to illuminate the film ( object ).
Note: The object is on the film or slide and the image ifs formed on the screen.
Slide or Film
The object is usually a film or slide whose image can be shown on the screen
The film is placed upside down in the projector. In this way, the image appears upright to the viewer.
However, the image still remains inverted.
Magnification of the Image on the screen:
M = v / u.
To obtain higher magnification; v must be increased and u must be reduced.
This can be done by:
i. Moving the screen away from projection lens.
ii. Moving projection lens nearer to the film object, or moving the film object nearer to the projection lens.
1. Light or Lens Camera
It is a light-tight box in which a convex lens forms a real image on the film.
The image formed is smaller than the real object.
Parts of the camera and their functions
Camera lens
It has a converging lens to focus the image from outside the camera.
It is used to form a small, real inverted image on a piece of photographic film.
The shutter
It controls the amount of light entering the camera by opening and closing the hole at different times
The closing and opening allows certain amount of light to enter through the hole
The amount of exposure depend on the amount of light available in high intensity the time is short compared
in dark
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Diaphragm
It controls the amount of light to enter the camera through the aperture. It opens when they is need of light
to take a picture.
Film
It act a screen
The film, which is normally kept in total darkness, contains a light sensitive chemical called silver bromide.
The film is found at the back of the camera and acts like a screen.
The camera has provision for adjusting the distance between the lens and the film.
Controlling light entering the Camera:
The diaphragm and shutter regulate the amount of light energy admitted through the lens.
1. Diaphragm : An adjustable ring of sliding plates called the diaphragm alters the diameter or aperture
of the hole through which light passes.
2. The Shutter : The shutter controls light entering the camera by opening and closing the hole at different
speeds.
Note:
1. Sometimes exposure time can be varied and are given in fractions of a second, e.g 1 / 1000 , 1 / 60 , e.t.c.
Fast moving objects require short exposures.
2. Exposure time is the time of opening the shutter to allow certain amount of light enters.
3. The shorter the exposure time, the smaller the amount of light incident on the film.
4. In bright sunlight, a short exposure time would be used. In dark conditions, a longer exposure time
would be used.
Focusing the Camera:
In simple cameras , the lens is fixed. In other cameras , the lens position is altered by turning the screw
mount of the lens or expandable box with accordion –like sides. This changes the lens –film distance i.e the
image distance.
THE EYES
The eye uses the converging lens as eye lens. The image if focused on the retina
FUNCTION OF DIFFERENT PART OF THE EYE
Iris
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It control the amount of light entering the eye. In bright light the iris draw closer to allow little light to pass
and in dark it widen up to allow more light to pass.
It control the size of the pupil like in bright right it make pupil to contract and became small and little light
can pass while in dark it widen up to allow more right to pass
The eye lens
It used to focus the image onto the retina
Pupil
It admit light it allow light to pass
Cornea
It protect the eye
It is transparent to allow light to pass through it
Retina
It is dark in colour found on the back of the eye to absorb light
It used as a screen where the image can be formed from the outside the eye
Optic nerve
It acts as the observer and carry the message to the brain for interpretation
Ciliary muscle
They are connect to eye lens so that they can change the shape of of the eye lens depending on the amount
on light entering
Accommodation of the eye
It the ability of the eye to focus a light from object placed at different position onto the retina and seen it
clearly.
This is done by changing the eye lens shape.
When the eye see a distance object, the ciliary muscle contract and the eye lens became thin and focus the
image onto the retina
When it want to focus the nearer object the ciliary muscle relax and the eye lens flattens to focus the image
on the retina
Defect of the eye
Short sight
It cause because the eye ball is too long and the light rays are focused before it reaches the retina
How to correct the defect
The defect can be corrected by using the diverging lens. The diverging lens diverge light light rays a little so
that they should meet at the retina
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Long sight
The image is formed behind the retina.
This is caused by relaxation of the eye lens making it to have long focal length.
To correct the situation convex lens should be used to help to converge light to enter the eye causing them to
meet the retina.
Comparison Between The Camera and Eye
Similarities:
1. Both have a converging lens
2. Both have their inside black
3. Both have a sensitive material where the image is formed like the eye has the retina and the camera has
the film
4. Both have a mechanism to control light entering. I.e. the eye has the iris and the camera has the shutter
and the diaphragm.
Differences:
1. Focal length change in eye while is fixed in the camera
2. Eye is normally always open the camera is always close expect the time taking picture
3. The eye has fixed image distance while camera image distance changes
Exercise
2003
1 (c)
i. What is the difference between a transverse wave and longitudinal wave?
ii. Give any two characteristics of a wave
2004
Question 5
a. Define the term oscillation in relation to swinging pendulum
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b. Figure 3 is a diagram of simple pendulum. The mass vibrate between point A and C through B
i. What happens to the speed of a mass as it moves from position
1A to B
2. B to C
ii. What is the speed of the mass at C?
iii. What happens to frequency and amplitude of oscillation of the pendulum as time increase
iv. State the energy changes of the mass as it changes from A to C
c. What happens to the frequency of the pendulum when the length of the string is changed
d. An object 2cm high is placed 7.5c in front of converging lens of focal length 5cm
i. Calculate the image distance
ii. Describe the nature of image
iii. Calculate the magnification of the image.
2005
Question 6
a. The figure below show the crest of straight ripples of water surface produced in a riplle tank by
generator
i. What kind of wave are represented by the crest
ii. What is the wave length of the ripples if there are 5 complete waves in a distance of 60cm
iii. What is the frequency of the ripples if four crest pass through point a in one second
b. What would happen if the waves moved from deep water to shallow water? Explain you answer?
c. Give a reason for answer
d. Describe constructive interference in water waves
e. What types of waves are radio waves
2006
Question 6
c. Figure 4 is a diagram showing a wave on a rope xy
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i.Calculate the wavelength
ii.If the rope xy is swung up and down 20 times in 2 seconds. Calculate the average speed of the wave
e. Figure 5 is a diagram showing to show water waves approaching two slits S1 and S2
i. Complete the diagram to show waves emerging on the other side of obstacle
ii. State two properties of the waves that demonstrated in the completed diagram
iii. What would happen to wave emerging on the other side of the obstacle if the width of S1 and
S2 were increased
2007
Question 2
a. Give factors that affect the frequency of vibrating spring
b. Figure 1 is a diagram of a wave
i. Name the part labeled A
ii. Calculate the frequency of the wave
iii. Calculate the velocity of the wave if its wave length is 50m
c. (i) State any three similarities between the camera and human eye
ii. What is the function of aperture in the camera?
d. An object is placed 15cm away from convex lens of focal length 10cm
i. Calculate the magnification of image
ii. What is the nature of image
2008
Question2
a. (i). Define the focal length of convex lens
ii. State two ways of determining the focal length of convex lens
iii. An object 10cm high is placed 25cm from the centre of convex lens of focal 10cm. Draw a
ray diagram to show the position of the image formed. ( Scale 1cm to represent 5cm )
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b. Figure show a diagram showing a mass hanging. If the mass is pulled to point A and released it
vibrates between point A and C through the rest position B
i Calculate the initial amplitude
iiAt which point does the vibrating mass have highest kinetic energy
iiiDescribe how the potential and kinetic energy changes as mass is vibrating from A to C
ivGive two reason why mass eventually stop vibrating
2009.
Question 2
a) Figure 2 show the part of the ray diagram
i. Complete the ray diagram to show the position of image
ii. Calculate magnification
b) Figure 3 is a diagram of the wave with a frequency of 2Hz
i. Name the type of the wave shown in figure 3
ii. Give any two properties of waves.
iii. What is the wave length of the wave?
c) Figure 4 is a diagram of a slide projector
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i. State the function of each of the following condenser lens, projector lens , can cave mirror
ii. Describe the nature of image formed
2010
Question 6
a. State any three difference between real image and virtual image
b. An object is placed 30cm in front of converging lens of focal length 10cm. Use the lens formula to
calculate
1. Image distance
2. Magnification
Question 7
a. State any two factors that affect the frequency of oscillating pendulum
b. Mention two characteristics of oscillating system
2011
Question 7
a. State the difference between transverse and longitudinal wave
b. Explain how focal length of a convex lens can be determined using a distance object
c. State any difference between the camera and human eye
Question 8
b. with an aid of labeled diagram explain the difference between waves passing through narrow and wide
gaps
2012
Question 4
a. Define oscillation
b. (i) What type of wave is produced by vibrating string
(ii). Calculate the frequency of a wave with a wave length of 2m and speed of 6m/s
c. State the difference between interference and diffraction of waves
Question 8.
a. With the aid of diagram explain how destructive interference in water waves occurs
b. Explain why waves refract when travelling from one medium to another
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CHEMICAL EQUATION AND REACTION
Chemical equation
It is the writing that represents what happens in chemical reaction
It is written inform of letter to show the reactant and products. The left hands side represents the reactant
while the left hand side is for products
Rules for writing chemical equations
1. Be sure that the reaction can take place as the experiment fact
2. Make sure you have written the correct formulae of reactants and products
3. Balance the number of atoms on the both side of the equation by multiplying molecules and adding
new atoms
4. In brackets after each reacts and products state the physical state like solid, gas. Or ,liquid
5. Never change the formulae of substance in order to balance it
Balancing equation
Write the equation using correct formulae for example
Na (s) + O
2
Na
2
O (s)
balance the atoms in both side
Na (s) + O
2
Na
2
O (s)
Number of the sodium in left side = 1 in the right side 2 , number of oxygen in the left hand side = 2 while in
the left is 1.
To balance multiply by 4 on sodium in left hand side and 2 in the right hand side to balance
4Na (s) + O2 2Na2O (s)
RELATIVE ATOMIC MASS (RAM)
It is the number of times an atom is heavier than a hydrogen atom or 1/ 12 the of a Carbon atom.
Relative Atomic Mass = Mass of 1 atom of element .no unit.
Mass of 1 atom of hydrogen
Relative molecular mass (RMM)
It is the sum of all relative atomic mass in a molecule?
Relative formula mass (RFM)
It is the sum of all relative atomic masses in molecular formula
Example :
Find the RFM for
1. Water ( H
2
O ) ;
2. Copper sulphate (CuSO
4
).
(RAM : Cu = 64 ; S = 32 ; O = 16 ; H =1 )
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Solutions:
1. RFM = 2 X ( 1 ) + 1 X ( 16 ) = 2 + 16 = 18
2. RFM = 1 X ( 64 ) + I X ( 32 ) + 4 X ( 16 = 160
THE MOLE CONCEPT AND AVOGADRO NUMBER ( L )
Moles
A mole is a collective term used to describe a quantity of material, just as a dozen is used .
One of substance contains approximately 6 x 1023 particles of that substance
This number is called Avogadro s number (L ) .
Mole calculations
Number of moles = Mass of Element in Grams
Relative Atomic Mass in Gram
N = m / RAM
Number of Moles in Molecule = Mass of the Substance
Relative Molecular Mass
It is conveniently arranged that 1 mole of atoms of any element has a mass equal to the Relative Molecular
mass but with units of g .
Example:
RAM of sodium Na =23. Therefore mass of 1 mole of sodium atoms = 23 g.
Example 1:
Calculate the number of moles in
i. 46 g sodium metal ( Na )
ii. 20 g sodium hydroxide ( NaOH ). (RAM: Na = 23 ;O =16 ;H = 1 ; cu = 64 ).
Solutions:
i. Number of moles = Mass / RAM
= 46 /23
= 2 .
ii. Number of moles = mass in g / RFM
= 20 /40
= 0.2
Example 2
How many grams are in?
a. 2 moles hydrogen
b. 0.2 mole copper ( RAM : H = 1 ; cu = 64 )
Solution
( a ) Mass = no. of moles x RFM
= 2 X 2 = 2 g.
( b ) Mass = no. of moles x RAM
= 0.2 X 64 = 13. 8 g
REACTION MASSES
Consider the chemical equation below
C + O
2
CO
2
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From the equation we can say the following
1. 1 atom of carbon react 1 molecule of oxygen to produce 1 molecule of Carbon dioxide.
2. 1 mole of carbon react with 1 mole of oxygen to produce 1 mole of Carbon dioxide
3. 12g of carbon will react with 32 g of oxygen to produce 44 g of carbon dioxide
Imagine we start with 6 gram of carbon dioxide. How many grams are required of oxygen and what will be
the mass of carbon dioxide?
From the equation
12 g of carbon react with 32 grams of oxygen
So 6 gram will react with x g ram of oxygen
32/12 = x/6
32x 6 / 12 = x
X = 16 grams
Example 1.
Methane burn in air to produce water according to the following equation
CH
4
+ O
2
CO
2
+ 2H
2
O
If 32g of methane was used up in burning
Calculate
a. Mass of oxygen required
b. The mass of water formed
From the balance equation
1 mole of methane reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.
Relative molecular of methane = 12 + 4 = 16
Relative molecular mass of carbon dioxide = 12+ 16x2 = 44
Relative molecular mass for oxygen = 16 x 2 = 32
Relative molecular mass of water = 16 + 2= 18. since they are two molecule = 2x 18 = 36
So 32g will react with y gram of oxygen
16/32 = 64/y
16y = 32x 64
y = 128g
Form the balance equation above
16 g of methane produces 36 gram of water
Therefore 32 g of methane will produce x gram of water
16/32 = 64/ x
16x = 32x36
= 72 g
Therefore 32 gram will react with 128 grams of oxygen and produce 72 g of carbon dioxide
Note:
Before doing calculation make sure that equation is well balanced
First calculate the RMM or RAM of reactants and products
Use the RMM and Ram to calculate the masses
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Reacting masses and balance equations
Reacting masses can be used to balance equation
Example:
A stream of chlorine gas was passed over and reacted with 2.24 of heated iron filings. The mass of chlorine
used up was 4.26 g. Work out the balance equation of the reaction.
(Hint Ram for iron is 56 and chlorine is 35.5)
Number of moles in iron = 2.24/56
= .04
Number of moles for chlorine = 4.26/ 35.5
= 0.12
0.12 moles of chlorine react with 0.04 moles of iron
Therefore 1 mole of iron will react with x moles of chlorine
0.4 = 0 .12
1 x
0.12 = x
0.04
3 moles = x
Therefore 1Mole of iron will react with 3moles of chlorine.
Fe + 3Cl Fcl3
Molar Volume and reacting masses
1 Mole of gas occupies 22.4dm3 at Standard Temperature and Pressure (STP) and 4.0 dm
3
at room
temperature and pressure. The temperature is 25 °C and pressure is 1atm.
Example1 :
Calculate the volume of oxygen gas O2 occupied by
(i) 2 moles ( ii ) 0.2 mole of the gas at room temperature and pressure
Solution:
( i ) volume of oxygen = no . of moles x 24 dm3
= 2 x 24 dm3
= 48 dm3
( ii ) volume = Number of moles x 24 dm3
= 0.4 x 24 dm3
= 9.6 dm3
Example2
Gaseous propane ( C
3
H
8
) born in oxygen to produce carbon dioxide and and gaseous water
i. Work out the balance equation of the reaction
ii. How many moles are there in 66g of methane?
iii. How many mole of oxygen is required to complete the combustion of 66g of propane
iv. If the same mass of propane are burnt in excess oxygen. What volume of carbon dioxide will be
produced at STP
v. If the same mass of propane are burnt in excess oxygen. What volume of water vapour will be
produced at RTP
Solution
1. Write down the balance equation as follows
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C
3
H
8
(g) + 5O
2
(g) 3CO
2
(g) + 4H
2
O (g)
Calculate the relative molecule masses as follows:
RMM of propane = 3x12 + 8 = 44
RMM of carbon dioxide = 3(12+ 2x16) = 132
RMM of oxygen = 5( 2x16) = 160
RMM of water = 4(2 + 16) = 72
Number of mole in 66 gram = m/ RMM
N = 66/44
= 3/2 = 1.5 moles
From the reaction equation.
1 mole of methane reacts with 5 Moles of oxygen
Therefore 1.5 moles will react with y mole of Oxygen
1/1.5 = 5/y
Y= 1. 5 x 5
y = 7.5 moles
1Mole of Methane produces 3 x 22.4 dm
3
of Carbon dioxide
1.5 moles will produce x dm
3
of Oxygen
X = 3x 22.4 dm
3
1
3x 22.4dm
3
= x
100.8 dm
3
= x
66g of methane produces 100.8 dm
3
1mole of methane produce 4 x 24dm
3
of water
1.5 moles will produce x dm
3
of water
1 = 4x 24
1.5 x
1.5x 4x 24dm3 = x
144 dm3 = x
66g will produce 144dm
3
of water if burn in excess
COMPOSITION CALCULATION
Step to follow
1. Work out the relative formula mass using Ram
2. Work out the masses of the elements in the compound from RFM
3. Express the percentage mass of elements X in the compound
Percentage mass of x = Mass of x in the compound x 100%
RFM of the compound
Example
Calculate the % composition of potassium nitrate
The formula is KNO3
K + N + 3O
RMM= 39 + 14 + 3(16) = 101
Percentage of K = 39 x100/101= 38. 62%
Percentage of N = 14 x 100/100 = 13. 86
Percentage of O = 48 x 100/101 = 47.52%
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EMPIRICAL FORMULA
It is a formula of a substance showing simplest ratio of atoms.
Example:
Calculate the empirical formula of organic compounds containing 92.3% carbon and 7.7 % hydrogen by
mass. (RAM: H = 1 ; C = 12 ) .
Since is by percentage by mass
It mean that carbon has a mass of 92.3 grams and hydrogen has a mass of 7.7 g
First calculate the number of moles of each element
Number of mole in carbon = 92.3/12
= 7.7 mole
Number of mole in 7.7 g of hydrogen = 7.7/ 1
= 7.7 mole
Find the ratio of atoms by dividing by the highest common factor of number of moles calculated
Ration of carbon = 7.7/7.7
= 1
Ration of hydrogen = 7.7/7.7
= 1
The empirical formula is CH
Or use the following method
C H
% by mass 92.3 7.7
in 100 g 92.3 g 7.7 g
Moles 92.3 g 7.7 g
1 2 1
= 7.7 7.7
Ratios of moles 1 1
Empirical formula = CH
CONCENTRATION
It is the amount of solute dissolved in a solvent.
Ways of expressing concentration
1. Mass per unit volume
Concentration = Mass of solutes
Volume of solution
This is expressed in g/l , g/cm
3
or g/cm
3
2. Mass per unit mass
Concentration = Mass of solute
Mass of solution (solvent)
It is expressed as solute per 100g of solvent
3. Moles per unit volume
Morality = Number of moles
Volume of solution in litres / dm
3
Percentage of solvent
It is expressed as percentage volume of solution. This is commonly used in beverage to show the percentage
of alcohol.
Concentration = Volume of solute (if liquid) x 100%
Volume of solution
Or concentration = Mass of solute x 100%
Mass of solution
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MOLARITY
Morality is a way of expressing concentration.
1 molar solution (1 M) is a solution prepared by dissolving 1 mole of a substance in 1 litre (dm3 ) of
distilled water .
Example:
Calculate the morality of solution prepared by dissolving 40 g sodium hydroxide ( NaOH ) in 500 ml of
water . (RAM: Na = 23 ; O = 16 ; H =1)
Solution:
Morality = Number of moles
Volume in dm3 .
Number of moles = mass of substance in g
RFM
But RFM = 1 X (23 ) + 1 X ( 16 ) +1 X ( 1 ) . = 40 .
Therefore number of moles = 40 g / 40 = 1.
Therefore morality = 1 mole / 500ml
= 1 / 0.5 dm
3
= 2 M .
STANDARD SOLUTION
Standard solution is a solution whose concentration is known.
Example 2 molar solution, a 5 molar solution, 0.1 molar solution .
Making standard solution
How to make a 1molar of sodium carbonate
1. Heat gently the solute you want to dissolve e. g sodium carbonate until all water evaporate . Left the
dish contain the sodium carbon in a dissector contain a dry agent silica gel.
2. Weigh the mass of the weight bottle and record it as mass 1.
3. Measure the mass of solute E.g. sodium carbonate by adding it using a spatula. The mass to be
measure should be the mass of weighting bottle plus the mass of solute. For example for 1M of
sodium 106 g should be measured plus the weight of the bottle.
4. Add enough distilled water to cover the particle of solute. Stir it with a glass stirring rod until all
solute dissolves.
5. Clean the stirring rod using distilled water and transfer the washing and the solution into 1000 cm3
volumetric flask using the funnel
6. Add distilled water until the mark on the neck. Use the washing from the washing bottle to top up the
solution level to the mark of the volumetric flask
7. A special stopper should be fitted on the mouth of the volumetric flask
HOW TO MAKE A SOLUTION FROM ANOTHER KNOWN SOLUTION
Example.
How to make 0.5 molar in 500cm
3
solution from 2 molar of NaOH
1. Using the formula M
1
x V
1
= M
2
V
2
. or C
1
V
1
= C
2
V
2
.
Calculate the required volume of the solution that will be diluted .
0.5 x 500 = 2 xy. Where y is the volume of solution to distilled
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125cm
3
= y
2. Using a measuring cylinder measure exactly 125cm
3
of 2 molar sodium hydroxide
3. Transfer the 2 molar sodium hydroxide to 500cm
3
volumetric flask.
4. Wash the measuring cylinder and transfer the washing to the volumetric flask using the funnel
5. Add the distilled water until it reaches the mark of 500cm
3
and take a cork and fit on it.
TITRATION
It is the gradual addition of a solution ( liquid ) to another solution ( liquid ) usually using a burette .
Application of titration:
The following are application of titration
1. To determine the concentration of unknown solution.
2. To make exact volume of a solution.
USING TITRATION TO DETERMINE CONCENTRATION OF UNKNOWN:
This can be used to determine the concentration of an acid or base.
A base or acid of known concentration is titrated to the solution whose concentration is not known until
neutralization takes place.
Phenolphthalein indicator is added either to the acid or base. At end point of titration (when a certain acid
reacts with the base), colour change takes place (from pink or purple to colourless or colourless to purple).
The volumes of the acid and base neutralized are used to calculate the concentration of the unknown.
Experiment: To find the concentration of hydrochloric acid from known solution by titration
Material
Burette
Conical flask
Measuring cylinder
Wash bottle
Dropper
Hydrochloric acid
1 Molar sodium hydroxide
Distilled water
Phenolphthalein indicator
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Procedures
1. Set the apparatus as below
2. Measure exactly 25 ml of Sodium hydroxide and put it in a conical flask.
3. Add few drops of phenolphthalein indicator
4. Add gradually an acid from the burette while you are shaking until it reaches the end point
when the colour just starts changing to colourless. Read the volume of the acid used
5. Repeat procedure 1 to 3 to obtain other values.
6. Add all the reading of the acid and find the average.
The following formula is used to calculate the concentration of the unknown acid:
Morality of acid (M
1
) x Volume used ( V
1
) = Morality of a base (M
2
) X Volume of a base ( V
2
).
Number of moles of acid
i.e . M
1
x V
1
= M
2
V
2
.
N
1
N
2
EXAMPLE:
Calculate the concentration of 20 ml Sodium hydroxide used to neutralize 40 ml of 0.2M
Hydrochloric acid.
Solution:
M
1
x V
1
= M
2
V
2
.
N
1
N
2
Balanced chemical equation for the neutralisation process is :
HCl ( aq ) + NaOH ( aq ) NaCl ( s ) + H
2
O ( l ).
Therefore N
1
=1 and N
2
= 1. Applying the equation we have:
0.2M x 40 ml = M2 x 20 ml . Since N1 = N2 =1
Therefore M
2
= 0.2M x 40 ml
20 ml
= 0.4M.
Other ways of finding concentration of unknown solution
Distillation
Simple distillation
The volume of the solution is measured later the solution is gently heated so that the solvent is
collected. The solute after distillation is dried and measure on triple beam balance. The mass of solute
is used to calculate the concentration of the solution
Fraction distillation
This method is used to find the concentration of liquid in liquid solution for example the concentration
of ethanol in water. The solution undergoes fractional distillation where by the solution with lowest
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boiling point is collected. Later it measured to find its volume and the concentration ids calculate
Evaporation Method
The volume of the solution is first measure and later the solution is gently heated until all solute
evaporate. Later the solutes are cooled and weighed to find its mass. The concentration is calculated
ad expressed as mass per volume or moles per litre
Freezing method
The volume of the solution is measure and later cooled until ice is formed. Later the solutes are dried
to obtain the solute which are later measure so that the concentration can be found
NB: Titration and evaporation are commonly methods used to find the concentration of solution
because it gives accurate results compare to distillation and freezing.
CHEMICAL REACTIONS:
A chemical reaction is the rearrangement of atoms to form a new substance or product.
A chemical reaction can be characterized by:
Change of colour.
Evolving of light
Evolving of heat
Production of sound.
Evolving of gases or fumes or steam.
TYPES OF CHEMICAL REACTIONS:
There are several types of reactions depending on different criteria used. E.g.
Displacement Reaction:
In this reaction an element or ion can be displaced by another element or ion.
Acid –base reaction:
In this reaction an acid is neutralized by a base to form a salt and water.
Oxidation –Reduction Reaction:
In this reaction one reactant is oxidized and the other reactant is reduced.
Exothermic Reaction:
It is a reaction in which heat energy is liberated and the temperature rises. In exothermic reactions ,
more energy is produced from the formation of new bonds than the energy required to break the
existing bonds.
Example.
Burning of carbon.
C ( s ) + O2( g ) CO2( g )
Endothermic Reaction:
It is a reaction in which heat energy is absorbed from the surroundings and the temperature falls.
In endothermic reactions, more energy is required to break old bonds than is liberated when new
bonds are formed.
Example:
Formation of hydrogen iodide from hydrogen and iodine.
H
2
( g ) + I
2
( l ) HI ( g )
Precipitation Reaction:
It is a reaction in which a precipitate (solid) is formed from two soluble solutions.
Example.
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Formation of silver iodide by mixing potassium iodide and silver nitrate solutions.
KI ( aq ) + AgNO
3
( aq ) AgI ( s ) + K NO
3
( aq )
Note:
1. For a chemical reaction to take place , there is braking of old bonds in reactants followed by
formation of new bonds in products
2. For bonds to be broken, energy has to be supplied.
3. When bonds are formed, energy is released.
HEATS OF REACTIONS:
Heat of reaction is the amount of heat released or gain when the 1 mole of substances reacts together
TYPE OF HEAT REACTION
1. Heat of combustion
2. Heat of Neutralisation
3. Heat of Formation.
Note:
The energy change that accompanies a chemical reaction is due to changes in chemical energy or
bonding energy between reactants and products
Heat of Formation:
It is the heat liberated or absorbed when I mole of a substance is formed from its constituent elements.
Heat of Combustion:
It is the heat liberated when 1 mole of a substance is completely burnt in excess oxygen.
Heat of Neutralisation:
It is the heat liberated when 1 mole of an acid reacts with 1 mole of bases
Enthalpy ( ∆ H )
It is the heat of reaction.
If the reaction is exothermic, ∆ H is negative.
If the reaction is endothermic, ∆ H is positive.
Heats of reaction can be categorized into the following
Exothermic and Endothermic heat of reaction
Exothermic heat of reaction
It the heat of reaction that cause the heating effect on the surrounding. The reaction gives off heat from
the system to the surrounding.
They always gain of heat by the surrounding so that the reactant can move from high energy to low
energy. The change in heat energy is always negative.
Energy Diagram for exothermic reaction
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∆ H = Final energy (energy of product) – initial energy (Energy of reactants)
Endothermic reaction
It is the change in physical or chemical which cause the cooling effect to the surrounding
The reactants take heat energy from the surrounding so that they can move from low energy level to
high energy level.
The energy is taken so that it is used to break bond and rearrange atoms in the reactants
The change in heat is always positive.
Energy diagram for endothermic reaction
NB The heat energy used to break bond is greater and is in excess, the reaction will release the extra
energy in form of heat hence exothermic but if the energy is small and extra energy is taken from the
surrounding and cause the drop in temperature in the surrounding the reaction ids endothermic.
Exothermic mainly happen in formation of bond because atom tend to loose extra energy to the
surrounding so that they can be in stable low energy state while endothermic happens in breaking of
bond because energy is required to break bonds what’s why they always energy drop in the surrounding
in endothermic reaction.
Heat Reaction Equation
For exothermic reaction the equation can be written in the following ways:
Writing the equation in full and the heat of reaction separate
2Mg + O
2
MgO (s) ΔH = -18kj
Or the heat of reaction written as part of equation
2Mg + O
2
MgO (s) + 18kj
In endothermic reaction the following ways can be used
By either writing the equation in full and the heat of reaction separate
NaCl(s) Na+ Cl- ΔH = +776kj
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or as part of the equation
776kj + NaCl(s) Na+ + Cl-
Experiment
Materials
2Test tube in a rack
Measuring cylinder
Thermometer
Spatula
Tap water
Ammonium chloride
Sodium hydroxide Procedure
1. Pour equal volume in each test tube approximately 5ml
2. Measure the initial temperature of water in each test-tube and record the results
3. Add half spatula of ammonium chloride in one and shake gently
4. Record the temperature of the ammonium chloride solution and record the results
5. Repeat step 3 and 4 using sodium hydroxide
Expected results
Solution
Initial temperature
Final temperature
Temperature change
(Final temperature- initial temperature)
Ammonium chloride
30
19
-11
Sodium hydroxide
30
49
19
Conclusion
Dissolving hydroxide is exothermic and release heat to the sounding while ammonium chloride is
endothermic and it get heat from the sounding
SPECIFIC HEAT CAPACITY
It is the amount of heat taken in or given out when 1 kg of substance change the temperature by
1degree Celsius.
S = H/MΔT
H = SMΔT
Example 1.
Find amount of heat need to raise the amount of temperature of 500g of water from 30 °C to 35 °C of
water. (Hint specific heat capacity of water = 4.2J/g-°C)
H = SMΔT
H = 4.2 x500 x 5
= 31500J
Calorific value
It is the amount of heat given out during the complete combustion of one gram of fuel
Example: During experiment, to determine the calorific value fuel a Student at certain Private
secondary school burnt 2g of Gawasha to find its calorific value. They find that it raise the temperature
by 15 degree of 100g of water. .
Find the amount of heat require to raise the temperature by 15 degree of 100gram
The calorific value of Gawasha
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Solution
H = SMΔT
= 100 x 4.2 x 15
= 6300J
The calorific value
If 2g produce 6300j of heat energy,
1 gram will produce 2/1 = 6300/x
= 6300/ 2
= 3150J /g
Finding out the Heat of Combustion /Calorific of an alcohol
1. Set the apparatus as below
2. Weigh the small spirit lamp containing the alcohol. But the lamp should be covered after the
weighing to prevent evaporation of the alcohol.
3. Put water of a certain mass in a metal can. Record the initial temperature of the water.
4. Light the lamp and put it underneath the can. The can should be heated until the temperature
rise to a certain level .E.g. about 60˚C.
5. Then re- weigh the spirit lamp to find the mass of the alcohol burnt.
Heat produced when the alcohol is burnt = Heat gained by the water.
But heat gained by water = mass of water in kg x temperature rise x specific heat capacity of water.
X g of alcohol produces Heat gained by water.
This has to be compared to the heat that can be produced by 1 mole of the alcohol mass of 1RFM of the
alcohol. E.g. For ethanol (C
2
H
5
OH )1 mole weighs 46 g.
ELECTRON TRANSFER
REDUCTION AND OXIDATION REACTION ( REDOX REACTIONS )
OXIDATION:
It is a reaction in which there is
1. Increase in oxygen content.
2. Loss of electrons
3. Increase of charge on the atom or ion.
Example of oxidation:
Burning magnesium in air, magnesium oxide is formed. In this cases magnesium gain oxygen and is
said to oxidized .
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2MgO(s ) +C uO(g ) 2MgO ( s ) + Cu.
Example 2. Zn˚(s ) + Cu
+2
( aq ) Zn
+2
( aq ) + Cu˚( s )
1. Zinc has lost electrons
2. Has increased its charge from 0 to +2
Therefore Zinc has been oxidized
From the equation we can say that
REDUCTION
Reduction is defined as
1. Gain in electron
2. Loss in oxygen
3. Decrease in oxidation number
For example
2CuO + C (s) 2 Cu0(s) + CO
2
(g)
From the equation
Copper has lost oxygen
Therefore cooper has be reduced
Example 2.
Zn˚(s) + Cu
+2
(aq) Zn
+2
(aq ) + Cu˚( s )
From the equation we can say
1. Copper has decreased its charge from +2 to 0
2. Copper has gained electrons
Therefore copper has being reduced
Redox reaction without oxygen
Consider the following equation
Zn˚(s ) + Cu
+2
( aq ) Zn
+2
( aq ) + Cu˚( s )
From the equation
Zinc has lost two electron and has also increase its charge from 0 to +2 there for it has been oxidized
Copper has gain two electron and it charge has decrease from +2 to 0 therefore it has been reduced by
gaining an electron
Oxidation number
It is the number given by the charge on a molecule or an atom.
For example Cu
+2
the oxidation number is +2
Rules of assigning oxidation numbers
1. All elements has oxidation number of zero when they are neutral
2. Oxygen has oxidation number -2 except when in gaseous state then its charge number is zero
3. Hydrogen has charge number of hydrogen is +1 except when it is an element
4. All molecules the total charge is zero
5. For charge ion the sum of the separate charges inside the ion must equal to final charge on its
self
Example
What is the oxidation number of S in SO
2
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From rule 2 oxygen has oxidation number of -2
Therefore 2 oxygen molecule has 2x-2 = 2
From rule 4 t the sum for a molecule is zero
Therefore S+ (-4) = 0
S= +4
The Charge for S is + 4
Example 2
Work out the oxidation number of MnO4-
From rule 5
MnO
4
-
Mn + 4( -2) = -1
Mn -8 = -1
Mn = +7
OXIDASING AND REDUCING AGENT
Oxidasing agent is a substance that cause another substance to oxidised
Oxidizing agent has the following properties
1. It gains electron
2. It loses in oxygen
3. It decreases its oxidation number
Reducing agent is the substance that causes another substance to be reduced.
The reducing agent has the following properties
1. It increase in oxygen content.
2. It gains oxygen
3. It increase its of charge
Example: In the following equations identify a reducing agent and oxidation agent
1. Zn˚( s ) + Cu+2( aq ) Zn
+2
( aq ) + Cu˚ ( s )
2. 2Mg˚ ( s ) + O
2
(g ) 2MgO( s ) .
3. Zn ˚ ( s ) + 2Ag +( aq ) Zn
+2
( aq ) + 2 Ag˚ ( s ) .
4. Cl
2
+ Br
-
2Cl
-
(aq) + Br
2
(aq)
Solution
1. Copper is oxidation agent because it has increase it charge while Zinc is reducing agent
because its charge has be reduced and has donated electrons
2. Magnesium is reducing agent because it has gain oxygen and has reduced oxygen to form
magnesium oxidised while oxygen is a oxidizing made magnesium to be oxidised
3. Zinc is reducing agent sagent because it has increase it charge while silver is oxidation agent
because its charge has be reduced
4. Bromine is oxidation agent because it has increase it charge while chlorine is oxidation agent
because its charge has be reduced
ELECTRON TRANSFER
DISPLACEMENT REACTION
It is a reaction whereby one atom or ion displaces the other .
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Consider the following reaction
When magnesium is placed in silver nitrate solution
1. Magnesium rod start to dissolve
2. The rod start to be cover with shiny silvery – black coatings
When zinc rod is placed in copper sulphate solution the following are observed
1. Zinc dissolve and a black brown coating of cooper is seen formed at zinc rod
2. The blueness of copper disappear after some time
When silver rod is placed in sodium chloride nothing happens
From the above observation
We can say
Magnesium displaces silver
Zinc displaced copper
Silver failed to displace sodium
From observation 1
Magnesium was dissolving according to equation
Mg° – 2e- Mg
+2
(aq)
The electron was transferred or handed over to silver as follow
Ag
+2
(aq) + 2e- Ag°(s)
Combining the two half equation
Mg° (s) – 2e- Mg
+2
(aq)
Ag
+2
(aq) + 2e- Ag°(s)
Overall equation
Mg° (s) +Ag
+2
(aq) Mg
+2
(aq) + Ag ° (s)
We can see the reason of formation of silver solid on magnesium rod
DISPLACEMENT TABLE OF METALS
It is a table with a list of metals showing how they displace each other in chemical reactions.
Metals at the top can displace metals lower down.
K Potassium
Na Sodium
Li Lithium
Ca Calcium
Mg Magnesium
Al Aluminum Direction of reactivity
Zn Zinc
Fe Iron
Sn Tin
Pb Lead
H Hydrogen
Cu Copper
Ag Silver
Experiment: To investigate the displacement reaction
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Materials:
Zinc metal
Copper metal.
Iron
Magnesium metal
Lead metal
Magnesium sulphate solution.
Copper sulphate solution
Lead sulphate solution
Iron sulphate solution
Zinc sulphate solution
4 Test tubes
Procedure
1. Fill each test tube with magnesium Sulphate and put all metal given in each test tube like.
2. Put each metal in each test tube expect magnesium metal record what happen in the table
below
3. Clean the metals with a sandpaper after use
4. Do the same with copper sulphate now exclude copper metal
5. Repeat the same with zinc sulphate, lead sulphate and iron sulphate.
Expected results
Magnesium
sulphate
Cupper
sulphate
Zinc sulphate
Lead sulphate
Iron sulphate
Magnesium metal
Disolve
Disolves
Disolves
Disolves
Copper metal
No reaction
No reaction
No reaction
No reaction
Lead metal
No reaction
Disolves
No reaction
No reaction
Iron metal
No reaction
Disolves
No reaction
Disolves
Zinc metal
No reaction
Disolves
Disolves
Disolves
The order of displacement will be as follows
Magnesium > Zinc > Iron > Lead > Cupper
ACTIVITY SERIES / ELECTROCHEMICAL SERIES
Any metal can either gain electrons from one metal or donate electrons to another metal. It show how
many electron a metal can lost or gain
K -e K
+
Na – e Na
+
Li – e Li
+
Ca –2 e Ca
2+
Mg –2 e Mg
2+
Al - 3e Al
3+
Zn -2e Zn
2+
Fe – 2e Fe
2+
Sn - e Sn
+
Pb - 2e Pb
2+
H
2
(g) –2 e 2H
+
Cu – 2e Cu
2+
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Ag – e Ag
+
OVERALL EQUATION OF DISPLACEMENT REACTION
It can be written by:
1. Writing down the half equation of the metal higher in the series
2. Writing down in reverse the half equation of the second lower metal in the series
3. Add up the two half equations.
EXAMPLE:
Placing zinc in silver nitrate solution:
Zn ˚( s ) - 2 e Zn
+2
( aq ) .
2Ag
+
( aq ) + 2 e- 2 Ag ˚ ( s ) .
Combining them we have:
Zn ˚( s ) - 2 e + 2Ag
+
( aq ) + 2 e- Zn
+2
( aq ) 2 Ag ˚ ( s )
Canceling electrons we have:
Zn˚ ( s ) + 2Ag +( aq ) Zn +2( aq ) + 2 Ag ˚ ( s ) .
COROSION / RUSTING
Corrosion is the process whereby metals dissolve and wear away in presence of oxygen and water.
The metal is oxidized as it loses electrons.
Oxygen acts as an oxidizing agent as it accepts electrons.
Most metals form oxides when they corrode.
But when iron is the metal that dissolves, the corrosion is called rusting.
Examples of rust:
i Pb – 2e- Pb
+2
(aq)
ii Fe - 2e- Fe
+2
(aq)
How rust is formed
Iron ionize according to
Fe – 2e- Fe
+2
(aq)
The electron are gained by water and oxygen
2H
2
O (l) + O
2
+ 4e- 4OH
-
(aq)
The hydroxyl ion (OH
-
) react with iron ions and water and air as follows
4Fe (OH)
2
(s) + 2H
2
+ O
2
(g) 4Fe (OH)
3
(s) (rust )
Conditions for rusting
1. Presence of water
2. Presence of oxygen from air
3. Presence of electrolyte e g sea water or acid solution
How to show that air and water cause rusting
1. Set the experiment as show below
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2. Put one nail in a test tube containing boiled water without air and label it as test tube A
3. Take another nail and put it calcium chloride powder to dry the air label it as test tube B
4. Take another nail and put it in tap water which contain air label it as test tube C.
5. Seal the three test tubes with wax to stop air entering the test tube.
Results
Test tube
Results
A with boiling water
No rusting
B with Calcium chloride
No rusting
C with tap water
Rusting take place
Conclusion
Rusting need air and water to occur.
Methods used to prevent rusting
1. Painting: This prevents water to react with the metal
2. By applying oil or grease―this prevents water to react with the metal
3. Electroplating i.e. covering a metal with another metal. This prevents iron from being attacked
by water and air.
4. Alloying, iron is mixed with another metal for example tin e.g. stainless steel. The metal to be
used for alloying commonly come from lower metals in the displacement series so that when
iron loses the electron the electron should be gained by the lower metal not water which
prevent rusting.
5. Sacrificial protection.
A magnesium ball is connected to an iron pipe. Iron does not dissolve; instead magnesium
dissolves and donates electrons to iron. Here magnesium is sacrificed in order to prevent iron
from rusting. Magnesium is higher than iron in the Activity series so it dissolves first and
donates electrons to iron. A large ball of magnesium or aluminum can protect an iron pipe many
kilometers long and for many years before being used up.
SIMPLE CELLS ( ELECTROCHEMICAL CELLS)
It is formed when two different metals are placed in electrolyte liquid and is joined by the external
wire. It cause the spontaneous redox reaction to occur when bringing current
Part of the electrochemical cell
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The simple cell has the following parts
Porous pot
1. It separate the two solution so that they should no general mixing
2. It allow current (ions) to pass from one solution to another completing the circuit
Metal container:
It is used to hold a solution plus it act as anode. It is made by the type of the solution of the metal that
will be displaced
The cathode rod:
It is made by the metal with is on top on the displacement series.
It is dipped in the solution containing the ions of the same metal.
It where oxidation happens, the anode dissolves
NB
Metals that are farther apart in the Activity Series or Voltage Series give higher voltage.
In the cell, a higher metal dissolves to release electrons and ions. Electrons can flow through the circuit
if metals are connected through a wire.
Polarization and Depolarization
Polarisation is the effects that cause the electric current to die away in a cell.
Cause of polarisation
In any electrolyte water also ionize as follows\
H
2
O H
+
+ OH
The anode attracts the hydrogen ion to be discharged. As a result there is formation of hydrogen gas
around the anode that act as an insulator preventing other ion to be discharged as a result it current dies
gradually
Depoloraisation
It is the process of trying out to avoid the cell no to die out
Potassium dichromate is used to as a depolarizer by reacting with hydrogen ion to form water.
The dry cell
Part of the dry cell and their function
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The solid carbon rod
Act as the anode of the cell. It receive electron from zinc plate
The zinc casing
It act the cathode which corrode and release electron to the external circuit
The Manganese dioxide Black powder:
Act as oxidation agent that act as depolarizer by converting hydrogen gas into water by adding oxygen
Moist Graphite granular
It helps manganese oxide to conduct electricity. The mixture is moist to allow ion to pass
Ammonium chloride Jelly
It allows the movement of ions and therefore passes the electric current through the cell
The brass cup
It act as a positive electrode
PROTON TRANSFER
LOWRY BRONSTED THEORY OF ACIDS AND BASES:
Lowry Bronsted theory of acid and bases tries to expand the meaning of acid and bases rather than the
ordinary meaning which say that acid and base are substance that produces water when they react. The
theory tries to explain why some substance behaves as acid although they do not produce water when
they undergo acid base reaction.
According to Lowry bronsted theory of acid and bases an acid is any substance that can donate or give
away proton awhile a base is any ion or a molecule that can accept a proton
In Lowry bronsted theory an acid is a proton donor. It provides protons or H
+
ions while a base is a
proton acceptor.
NB: The hydrogen ion is simply a proton. This is because a hydrogen atom contains a proton and an
electron. As H
+
, the electron is removed. The acid can act as a proton donor if there is a base (proton
acceptor ).
Example of Lowry Bronsted acid and base reaction
HCl ( g ) + NH
3
( g ) NH
4
( s ).
HCl is donating H
+
to NH
3
Therefore HCl is L/B acid is a proton donor
NH
3
is accepting H
+
to form NH
4
+
The ammonium is a L/B base because it has accept a proton from HCl
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NH4
+
+ Cl
-
HCl + NH
3
From the equation
NH
4
+
is behaving as an acid while Cl
-
behaving as a base
IONIC EQUATION:
It is an equation which shows the ion which causes acidity H
+
( aq ) reacting with the ion which causes
alkalinity OH
-
( aq ) to produce neutral water.
The equation summarises the reaction between any acid and alkali in aqueous solution.
Eg.
HCl ( aq ) + NaOH (aq ) NaCl ( aq ) + H
2
O( l ).
H
+
( aq )- ( aq ) + Na
+
( aq + OH
-
( aq ) Na
+
( aq )+ Cl
-
( aq )+ H
2
O( l ).
By taking out spectator ions: we have;
H
+
( aq ) + OH
-
( aq ) H
2
( l )
NB. Spectator ions here are those which are unchanged on either side.
WATER AND LOWRY BRONSTED SYSTEM
Water as base
Water molecule can accept a proton: e.g.
HCl ( aq ) + H
2
O( l ) H
3
O
+
( aq) + Cl
-
( aq ).
The water molecule accept hydrogen proton to form a H
3
O
+
is a hydronium ion or oxonium ion.
During the reaction water tend to accept a proton from an acid. In this situation it act as a L/B base
HNO
3
(aq) + H
2
O H
3
+(aq)+ NO
3
-(aq)
H
2
SO
4
(aq) + H
2
O H
3
+(aq) +HSO
4
- (aq)
Water as acid
Water can donate a proton. Eg.
H
2
O( l ) +NH
3
( aq ) NH
4
+
( aq ) + OH
-
( aq ).
H
2
O + H
2
O H
3
O
+
(aq)+ OH
-
( aq ).
NB 1 A reaction with water is called hydrolysis.
2. Substance which can behave both as L/B acid and L/B base is called Amphoteric.
CONJUGATE ACIDS AND BASES:
These are pairs of acids and bases.
CONJUGATE BASE
To find a conjugate base subtract H
+
from its conjugate acid.
Example
Conjugate acid Conjugate base
H
2
SO
4
- H
+
HSO
4
-
HCl - H
+
Cl
-
H
3
O+ - H
+
H
2
O
CH
3
COOH - H
+
CH3COO-
NH
4
+ - H
+
NH
3
To find a conjugate acid add H
+
the conjugate base.
In an L/B ACID-BASE EQUATION, the conjugate acid and its conjugate base are on opposite sides.
STRENGTH OF AN ACID AND A BASE:
Strong acids are acids which completely ionize when dissolved in water. Examples are:
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1. Hydrochloric acid ( HCl ).
2. Sulphuric acid ( H
2
SO
4
)
Weak acids are acids which do not completely ionize when dissolved in water. Some of the molecules
remain un-ionized. Examples are:
1. Ethanoic acid (CH
3
COOH).
2. 2. Carbonic acid ( H
2
CO
3
)
3. Sulphurous acid ( HSO
4
-
)
4. Hydrofluoric acid (HF )
A strong base is a substance which ionizes completely when dissolved in water.
Examples are:
1. Sodium hydroxide (NaOH).
2. Potassium hydroxide (KOH).
A weak base is a substance that does not ionize completely when dissolved in water. Examples are:
1. Ammonia ( NH
3
). 2. Ammonium hydroxide ( NH
4
OH).
NB 1. For a strong acid or base, there is a single arrow in the ionic equation. For instance,
i. HCl ionizes as follows:
HCl ( g ) water H
+
( aq ) +Cl
-
( aq ) .
ii. NaOH ionizes as follows:
NaOH ( aq ) water Na
+
( aq ) + OH
-
( aq ) .
2. For a weak acid or base, there are two arrows in the ionic equation. Because there is formation of
reactants and products at the same time. For instance;
Ethanoic acid ionizes as follows:
CH
3
COOH H
+
( aq) + CH
3
COO
-
( aq )
ii Ammonium hydroxide ionizes as follows :
NH
4
OH NH
4
( aq ) + OH
-
( aq )
DIFFERENCE BETWEEN STRENGTH AND CONCENTRATION OF AN ACID
The strength of an acid tells how easily it ionizes (dissociates) to produce hydrogen ions.
The concentration of an acid indicates the proportions of water and acid present in aqueous solution.
NB 1. A strong acid is still strong even when it is in dilute solution.
2. A weak acid is still weak even when it is concentrated.
DETERMING THE STRENGTH OF AN ACID OR BASE:
1. BY USING CONDUCTIVITY APPARATUS.
By passing electricity through the acid or base. The size of current can be higher if the acid or base is
stronger. The size of current can be lower if the acid or base is weaker.
2. BY USING PSCALE:
1. UNIVERSAL INDICATOR.
This can be used to find the pH value of the acid or base.
Few drops of the indicator can be added to the solution and there can be a colour change
The colour of the reaction mixture can be matched with the colour on the chart of the indicator. The pH
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value can read on the chart.
2. The pH Meter can also be used to find the pH value of the acid or base.
When its two electrodes are dipped in the solution, the Ph of the solution value can be shown.
IONS IN SOLUTION
The conductivity circuits
Conductivity is the ability of a substance to allow electric current to pas through it.
Uses of the conductivity circuits
1. To show that a current flows in all parts of the circuits
2. To find the kind of electrolyte present in a beaker from the the reading of the ammeter.
3. as apparatus of electrolysis
Parts of the conductivity circuits
Electrodes.
It is a path where electric current enter and leave the solution.
It is made of nonreactive metal like graphite, platinum
They are two electrolyte, the positive electrode call the anode which is connect to the positive terminal
of the cell and the negative electrode which is called the cathode and is connect to the negative terminal
of the cell
The anode attract the negatively charged ions called the anions while the cathode attract the positive
charged ions called the cations.
Electrolyte
It is the substance that will conduct electric current by mean of ions in liquid form.
It is usually made by the substance that conduct electricity in molten form
Type of electrolytes
Strong electrolytes
Large current pass through the solution because it has many ions present in the solution
Strong electrolyte dissociate complete in water when dissolved. They complete dissolve in water.
Weak electrolyte
They partially allow current to flow through the conductivity circuits. They have few ions that carry
electric current. They do not completely dissociate in water and are partly soluble in water.
Non electrolyte
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The do no conduct electricity because they have no ions that can carry electric current through the
conductivity circuit. They do not dissociate themselves in water and are insoluble in water.
ELECTROLYSIS
It is a process whereby current passes through the liquid or solution and breaks down the liquid into its
individual components.
ELECTROLYSIS OF SOLUTION
All metals ions are discharged at the cathode.
All Non-metals are produced at the anode except hydrogen ion .
NB: Reactive metals are not formed at the cathode, except electrolysis of sodium chloride using a
mercury cathode
Factors that affect the electrolysis of a substance
1. Amount of current: ions are discharged when current is high. Compared to low current
2. Type of the Electrolyte: Strong electrolyte are discharge easily and fast compare to week
electrolytes
3. Depth of electrode: if the electrode are dipped deep in the solution they provide a large
surface are for the ions to be discharged compare to small and shallow dipped electrodes
4. Concentration of the solution: Very concentrated solution provides enough ions that will
be discharged compared to dilute solution.
5. Distance between the electrodes. The electrodes are far from one another , ions travels
long distance to be discharged compared to electrode that are closer to each other.
Electrolysis of copper sulphate using copper electrodes
At the node
The anode dissolves.
This mean that electron are given into the solution
Cu˚(s) – 2e- Cu
2+
(aq)
At cathode
Copper is deposited
This means that copper solids are formed
Cu
2+
+ 2e- Cu ˚(s)
The colour of the solution remain the same because anode dissolves into solution maintaining the
colour of the solution
Preferential discharge of ions at the cathode
When sodium is dissolves in water
There are two ions in the solution Na
+
and Cl
-
But during electrolysis of sodium chloride
Hydrogen gas is released from the cathode not sodium
This means that they are two ions that are attracted towards the cathode, sodium ion and hydrogen ions
In aqueous solution water dissociate itself as follows
H2 H++ OH-(aq)
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Therefore H+ions are attracted toward to cathode to be discharge because they need H+ need lowest
energy to be discharged compared sodium. Hydrogen ion is lighter compared to sodium ion and travel
to the cathode to be discharged.
Factor that affect preferential discharge
the following factors can decide which one should be discharged first
1. the ion should have lowest energy for discharge
2. the ion should be in high concentration
3. the nature of the electrode. If the anode is made of the material like the electrolyte it will be
preferred to be discharged.
NB. If the concentrations of the negative ions in solution are approximately the same, the order of
discharging is :
OH- aq
I-
Br- aq Ease of discharging decreases going down
Cl- aq
NO3- aq
SO4-2 aq
USES OF ELECTROLYSIS
1 Electroplating
2.Purification of copper
3. Extraction of metals such as sodium, aluminum
4. Manufacture of sodium hydroxide through electrolysis of sodium chloride either using the
diaphragm cell or the Kellner Solvary cell. In both cases , the products are sodium hydroxide, hydrogen
and chlorine.
ELECTROPLATING:
It is a process of forming a very thin layer coating of a metal on the surface of another metal using
electrolysis.
Reason for electroplating materials.
1. T o prevent rusting or corrosion
2. For decorative purposes.
3.To repair worn out machinery.
CONDITIONS FOR ELECTROPLATING:
1. The metal to be coated is taken as a cathode.
2. The covering metal is made the anode.
3. The electrolyte should be a soluble salt of the metal used to cover another metal.
NB. For effective plating , the cathode must be clean and free from grease.
Example of electroplating
Coating copper with zinc.
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Zinc dissolves. The zinc ions move towards the ( Cu ) and they gain electrons to become zinc atoms.
Zn0( s ) - _______ Zn+2aq : at anode : Zinc dissolving
Zn+2aq + 2e- _______ Zn0( s ) : at cathode : zinc coating copper.
.PURIFICATION OF COPPER:
the impure copper is made the anode while the pure copper is made the cathode . The impure copper
dissolves. The pure copper is formed art the node as shown in the diagram
Cu0( s ) –2e- ______ Cu+2aq : on anode.
Cu+2aq + –2e- _______ Cu0( s) : on cathode
Exercise
2003
Question 3
a. (i). What is a mole?
ii. a solution was made by dissolving 8 g of sodium hydroxide in 100cm3 of water. Calculate
the morality of the solution (RAM: Na = 23, O = 16 H = 1)
b. Copper ( Cu) reacts with silver ions (Ag
+
) according to the following equations
Cu ˚ ( s ) + 2Ag +( aq ) Cu
+2
( aq ) + 2 Ag˚ ( s ) .
i. What is the meaning of zero (o) sign on Cu ˚
ii. Pick out the oxidizing and reducing agent from the equation
iii. Write the half equation of the reaction
6d. (i). What is meant by empirical formula?
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iii. Work out the empirical formula of the compound that has the following percentage
composition by mass of element ( C = 40 %, H = 6.67% and O = 53.33% ) RAM : H = 1 O
= 16 and C = 12
Question 8
a. What is the difference between an exothermic reaction and endothermic reaction?
b. Given that the reaction between methane and oxygen to produce carbon dioxide and water is
exothermic and dissolving of ammonium nitrate is endothermic. Draw energy level diagram to
illustrate this
c. Describe how 250cm
3
of 1 molar cooper sulphate could be prepared using hydrated copper
sulphate crystals ( CuSO
4
5H
2
O) ( the molar mass of CuSO
4
5H
2
O is 250)
2004
Question 7
a. Describe how the concentration of 20cm
3
of sodium chloride can be determined using
evaporation method
b. Suggest the sources of errors
c. 100cm
3
solution of concentration 20g/l is diluted by raising the volume to 250cm
3
with distilled
water. Work out the concentration of new solution.
2005
Question 7
a. Draw the diagram that could be used to electroplate an iron nail with cooper using copper
chloride
b. Explain what happens during the process of electroplating of the iron nail in (a). Support the
explanations with relevant chemical equations.
Question 4 PPII
a. Define the following terms
i. Oxidation
ii. Reduction
b. With an aid well labeled diagram , describe the experiment you would carry to show both air
and water are necessary for rusting
2006
Question 5
a. (i) . Name the ion responsible for acid condition properties in a substance
ii. Why is carbonic acid a weak acid while hydrochloric acid a strong acid?
b. Calculate the volume of 0.1M sodium hydroxide that is needed to neutralize 20cm
3
of 0.1
hydrochloric acid
c. (i). What does the symbol mean in a chemical equation
iii. Complete the following chemical equations
1. HSO
4
-
(aq) + __________ H
3
O
+
(aq) ______________ ( aq)
2. H
3
PO
4
(aq) + __________ __________ (aq) + H
2
PO
4
-
(aq)
d. The following is the part of activity series
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Lithium (Li)
Sodium (Na)
Magnesium ( Mg)
Lead ( Pb) Increasing reactivity
Hydrogen ( H)
Copper ( Cu)
Silver (Ag)
i. State whether copper (Cu) will react with a solution of magnesium sulphate ( MgSO
4
)
ii. Explain your answer
iii. Which element is most reactive
iv. Give a reason for you answer
e. (i). Write the half equation for the reaction between silver nitrate (AgNO
3
) and sodium (Na)
ii. Name the reducing and oxidizing agent
2007
Question 3
a. Ammonium is example of strong bases
i. What is a strong base?
ii. Write a chemical equation to show the ionization of ammonium in water
i. Identify conjugate acid base pair from the equation in 3 a.ii.
b. (i) Draw the energy level diagram for the following chemical reaction
NaOH (aq) + HCl (aq) NaCl + H
2
O (l) ∆H = - 57Kj/ Mol.
i. Is the reaction in 7 b ii. Endothermic or exothermic?
ii. Give a reason for your answer
iii. What is the meaning of (aq) and (l) in the equation?
c. A 300mg tablet was completely dissolved in 10ml of water , If the molecular formula of the drug
is C
9
H
8
O
4
i. Calculate the number of moles in the tablet
ii. Calculate the concentration of the solution
6e. the following ids a part of reactivity series
Magnesium ( Mg)
Aluminum (Al) Increasing reactivity
Zinc ( Zn )
Iron (Fe)
Copper ( Cu)
i. Which of the two elements will displace zinc (Zn) from its oxides?
ii. Give a reason for your answer
f. The following are half equation for the reaction between magnesium ( Mg) and Silver nitrate (
AgNO
3
)
2Ag
+
(aq) + 2e 2 Ag˚ (s)
Mg˚(s) – 2 e Mg
2+
(aq)
i. Write the chemical equation for the reaction
ii. Name the reducing and oxidizing agent
2008
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Question 7
b. What is the difference between oxidation and reduction in term of electron transfer
c. Explain how each of the following prevent rusting in iron
i. Painting
ii. Galvanizing
3PPII
a. Define a standard solution
b. Explain how 500cm
3
sodium chloride solution can be prepared using sodium chloride
solution crystals. The explain should include al then necessary mathematical calculation. (
relative atomic mass of Na = 23 and Cl = 35 )
2009
5 f. Silver nitrate (Ag
+
react with Fe according the following equation
Fe ˚ ( s ) + 2Ag
+
( aq ) Fe
+2
( aq ) + 2 Ag˚ ( s ) .
i. What is the meaning of the symbol 2+ on iron
ii. What is the oxidation number of silver before reaction
iii. Which substance has been reduced?
iv. Give a reason for your answer in 5f iii. Above
Question 8
a. What is a electrolyte
b. With the aid of the diagram describe an experiment you can carried out to compare
electrical resistance of potassium nitrate solution and potassium chloride solution.
2010
Question 2
a. State two advantages of electroplating a metal
b. Below is a part of displacement series metals
Calcium (Ca)
Magnesium ( Mg)
Aluminum (Al) Increasing reactivity
Zinc ( Zn )
Iron (Fe)
Copper ( Cu)
i. Which metal would displace all others metals from their solution
ii. Give a reason for your answer
iii. What would happen if magnesium metal was placed in copper sulphate solution
c. Water react with carbonate ion according to the following equation
H
2
O (l) + CO
3
2-
OH
-
(aq) + HCO
3
-
(aq)
Give one conjugate acid base pair in the reaction
d. Define oxidation in term of oxygen content in a substance
e. The chemical equation below shows the displacement reaction between zinc ( Zn) and lead
nitrate (Pb(NO
3
)
2
)
Zn (s) + Pb(NO
3
)
2
Zn(NO
3
)
2
+ Pb
Name the reducing and oxidation agent in the reaction
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f. Figure 2 is a diagram showing a set of experiment to investigate conditions for rusting of
iron
i. Why was anhydrous calcium chloride used in test tube Y
ii. Why was water5 in test tube Z boiled?
iii. State any two condition necessary for rusting
2011
Question 3
a.(i). Define Molar volume of a gas
ii. Calculate number of mole of sulphur dioxide occupying 120dm
3
at room temperature and pressure
b. (i) define acid according to Lowry Bronsted theory
ii. Describe how a hydronium (H
3
O
+
) ion is formed
c. Explain why bond breaking is endothermic while bond making is exothermic
d. (i) Mention any two ways of preventing corrosion in metals
(ii). Calculate the oxidation number of nitrogen (N) in nitrate (NO
3
-
) give n that the
oxidation number of oxygen (O) is – 2
2012
Question 5
a. Define electroplating
b. Iron ( Fe) displaces copper (Cu) from copper sulphate solution (CUSO
4
)
i. Write down a balanced equation for the reaction
ii. What is the reducing agent in the reaction?
iii. Give a reason for your answer
c. What is the difference between oxidation and reduction?
d. (i). define concentration of a solution
ii. The volume of sodium hydroxide solution (NaOH) of concentration 20g/l is increased
from 60Cm
3
to 600cm
3
by adding distilled water. Calculate the concentration of the
new solution in g/l
e. Figure 3 is an energy level diagram for the reaction between magnesium (Mg) and oxygen
gas (O
2
)
i. Is the reaction exothermic or endothermic
ii. Give a reason for your answer
iii. State the meaning of the arrow in the diagram
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ELECTRICITY, MAGNETSIM AND ELECTROMAGNETIC
INDUCTION
Electricity is flow of electrons the negative
TYPES OF ELECTRICITY
1. Static Electricity 2. Current Electricity
Static Electricity:
It is electricity due to static charges. Examples
i. Rubbing plastic ii. Electricity in a capacitor.
Electrostatics:
It is static electricity or stationary electric charges.
An object becomes electrically charged if it has too many or few electrons.
WAY OF MAKING STATIC ELECTRICITY
1. By contact
The charged material is brought close to uncharged material. The charged material come into
contact with uncharged object
2. By induction
The charged material are is brought close to uncharged material. The charged object is just
brought closer to un charge object.
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3. By friction or by rubbing
Uncharged material is rubber by a polythene material by the hair
Electric Field
Static charges create a field of force in the space around them. This field is called Electric field. The
electric field can be mapped using bits of hair , just as magnetic field can be mapped with iron fillings.
Static charges obey the inverse square law: i.e.
The force between them becomes 1 /4
th
as much if they are moved twice as far apart.
Electric charges can be detected using a device called Electroscope.
APPLICATIONS OF ELECTROSTATICS
1. They are used in electrostatic filters to clean air. The smoke pass through the negatively charged
metal plate where they attain the negative charge and are attracted towards the positively charged
plate leaving the smoke free of particles.
2. The are used in some photocopiers , light reflected from the white parts of the paper cause the
charge on the drum to disappear from corresponding parts of the drum which later attracts the toner.
3. Ink jet printer, ink is passed through oppositely charged particles and the amount of reflection
depends on charge and Pd across the plates.
4. They are used in paint –spraying: The car body is given the negative charge while the positive paint
is give positive charge. This ensures even distribution of paint on the body of the car.
5. They are applied in electrostatic loud-speakers: Charged air molecules vibrate directly by
electrostatic forces.
6. They are important in the operation of many electronic devices such as : Capacitors; cathode –ray
tube ; which are used in radar, television and electrical test equipment
DANGERS OF ELECTROSTATICS
1. It cause lighting which cause the lost of properties and life of people.
2. Refueling. It can cause fire when the body of the vehicle or aircrafts charged by rubbing in the air.
3. Operating theatres - charged object attract germ and dust into the body of the patient during
operation.
4. Computers- they are easily damaged by electrostatic
CURRENT ELECTRICITY
It is the electricity that flows due to flow of electrons.
Sources:
Cells ( batteries ) – solar panel – electric generators – thermocouple.
Some terms used in Current Electricity:
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Electric Current ( I )
It is the flow of electrons in a complete round path called circuit.
It is measure in units called Amperes ( A ).
It is measured using an instrument called an Ammeter.
Voltage ( V )
It is the electromotive force used to drive electrons in a circuit.
It is also the potential difference drop across an element.
It is measured in units called volts ( v ).
It is measured using an instrument called a voltmeter
Electrical resistance ( R )
It is the opposition of a conductor to current.
It is also the opposition to the flow of electrons in a circuit.
Factors Affecting Resistance of Conductor:
1. Length of a conductor:
As the length increases, resistance increases. The electrons travel a long way during the process they
bump into many atoms.
2. Cross –section area.
As the cross-section area decreases, resistance increases,
The passage for the flow of electrons becomes less so electrons bump into many atoms.
3. Temperature of the conductor:
As the temperature increases, resistance decreases. The kinetic energy of the atoms increases the
vibration of atom like atoms tend to vibrate with large amplitude which block electrons. But in
semiconductors the resistance decreases at temperature increases because the kinetic energy by electron
so that it jump from valence band to conduction band.
4. Nature of the material:
Naturally, some materials have high resistance while other materials have lower resistance. For
example: Silver is a best conductor, but copper is the next best and nichrome has higher resistance.
Expriment1
The effects of length on resistance of the wire
Set the circuit as below
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Connect the wire between AB . Note the reading of the ammeter and the voltmeter.
First connect 100cm wire and note the reading of the ammeter
Repeating by measuring the reading of ammeter and voltmeter with 80cm, 60cm.40cm and then 20cm
NB the same circuit can be used to verify with difference cross section area
Results
Length
Voltage (V)
Current (I)
Resistance (V/I)
100
3.3
0.47
7
80
3.1
0.52
6
60
2.8
0.56
5
40
2.4
0.6
4
20
2.1
0.7
3
From the result we can see that length is directly proportional to the resistance of the wire
like
R L
R = kL, k = R/L
if there two length of the same wire.
Then R
1
= R
2
L
1
L
2
Example:A 20cm wire has Resistance of 3ohms . Find the resistance of the same wire if the length is
70cm.
Given : R
1
= 3 , L
1
= 20cm , L
2
= 70cm., R
2
= ?
R
1
= R
2
L
1
= L
2
3 = R
2
20 70
(3x70) = R
2
20
10.5 ohms = R
2
From experiment it show that resistance directly proportional to the square of diameter of the wire.
R d
2
R = kd
2
if the are two wire of the same type then
R
1
= R
2
d
2
1
d
2
2
Example
Find the resistance of nichrome wire with a diameter of 3mm if a 2mm wire ha resistance of 5ohms
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R
1
= R
2
d
2
1
d
2
2
5 = R
2
2
2
3
2
5 x 9 = R
2
4
11.25ohms =R
2
OHM S LAW:
It state that voltage is directly proportional to current
Mathematically
V I
V = kI
k is proportionality constant called resistance ®
therefore
V = RI
R = V / I
Resistance ( R ) is measured in units called ohms(Ω).
RESISTORS:
These are conductors or electronic devices intended to have resistance.
Some resistors are made either from wires of special alloys or from carbon.
FINDING VALUES OF RESITORS:
Resistance of resistors can be found using Colour code and Resistance code.
Resistors colour code
It is the method used to indicate the resistance of a resistor . Chosen colours are used to represent the
numerical values
The resistor has colored bands or rings printed on it. Each band or rings are printed on a resistor, each
representing a number.
The first band is for the 1
st
digit; -
The second band is for the second digit.
The third band is for the number of zeros.
The fourth band or absence of a band is for tolerance .There is an allowance that resistance can vary
from the marked value.
Colour
Tolerance
Brown
1.00%
Red
2.00%
Gold
5.00%
Silver
10.00%
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Example:
No colour
20.00%
Colour
Number
Black
0
Brown
1
Red
2
Orange
3
Yellow
4
Green
5
Blue
6
Violet
7
Gray
8
White
9
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Find the resistance of a resistor whose colour code is Red-yellow- orange silver.
Solution
Red = 2 , yellow 4 and silver is 10% tolerance
Therefore
Resistance is 24000
Tolerance = 10/100 x 24000
= 2400Ω
Therefore Resistance = 24000- 2400 = 21600 OR
= 24000 + 2400 =26400.Ω
The true value of resistance is between 26400Ω and 21600 Ω
Resistance Code:
Resistance values are printed on the resistor using letters.
Letters R and K indicate the position of a decimal point in the resistance value. As shown below;
Code Resistance Value
R45 0.45
5R4 5.4
5K4 5.4 k
22K 22k
Tolerance for the Resistance Code:
Letter Tolerance Letter Tolerance
F 1% K 10%
G 2% M 20%
J 5%
Example1
example 2
The resistance of the resistor is 4700 ohms or 4.7 kilo ohms
RESISTANCE IN CIRCUITS
Resistors in series
Resistance is placed end to end in a straight line.Resistance increases when resistors are connected
in series. This because each resistor, one after the other, resists the electrons as there is only one
conducting path.
Investigation
Investigating the behavior of resistors connected in series
Materials
3 carbon resistors
2 cells
Switch
9 connecting wires
Ammeter
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Voltmeter
Procedure
1. Measure the resistance of each resistor and find the sum
2. Connect the three resistors as below
3. Note the ammeter and voltmeter reading and record it
4. Divide the voltage by current to find the resistance
5. Compare the two results
OBSERVATION: From the experiment it has been seen that the sum of individual resistor is
equal to the total resistance in series.
How to find total resistance in series
From Junior certificate physical science you learn that current is the same at any point in series
circuit. You also learn that voltage from the cell is equal to the sum of voltages in each components
in series circuit.
Like V
s
= V
1
+ V
2
+ V
3
From Ohm law we know that V= RI
Then substituting V for RI we get
R
s
I =R
1
I + R
2
I + R
3
I
diving teach term by I since it is the same in each component we get
Rs = R
1
+ R
2
+ R
3
We can conclude that resistance or resistors in series is additive. i.e.
Rs = R
1
+ R
2
+R
3
+- - - - + R
n
. where n = the n
th
resistor .
Example. Find total resistance of the resistors in the circuit below.
Rs = (3 + 5 + 9) ohms
Rs = 17ohmResistors in Parallel:
Each resistors are connected in each junction or each conducting path .
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Resistance decreases when resistors are connected in parallel because is that in a parallel circuit
there are more conducting paths so more electrons flow per unit time.
Investigation: The behavior of resistors in parallel
Materials
3 carbon resistors
2 cells
Switch
9 connecting wires
Ammeter
Voltmeter
Procedure
1. Measure the resistance of each resistor
2. Calculate the reciprocals of each resistance and find the sum
3. Connect the three resistors as below
4. Note the ammeter and voltmeter reading and record it
5. Divide the voltage by current to find the resistance
6. Compare the two results
Observation: You have observed that the total resistance in parallel is equal to sum of
reciprocals of all resistance connected in the circuit
From Junior certificate physical science you learn that current arriving at the junction is equal to the
current leaving the junction in other words current from the cell is equal to sum of currents in
junctions. You also learn that voltage is the same in each component connected in parallel circuit.
I
s
= I
1
+
I
3
+ I
3
From ohms law I = V/R
substituting V/R for I
VV VV
R
s
= R
1
+ R
2
+ R
3
Dividing each term by V
V V V V
VR
S
= VR
1
+ VR
2
+ VR
3
1/R
s
= 1/ R
1
+ 1/R
2
+1/ R
3
To find the total resistance in parallel circuit add the reciprocals of each resistor in parallel.
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For example, Find the total resistance for the resistors in the diagram
The total resistance R
t
=1/R1 + 1/R2 +1/R3
1/ R = 1/8 + 1/6 + ½
= 3+ 4 + 12
2 4
19/24
R = 24/19
= 1.263 OHMS
example 2
A certain student connected 3 ohm, 5 ohm and 6 ohm in parallel circuit. She later add two resistance
with values of 7 and 8 ohm in series circuit. Find the total resistance in the circuit
Solutions
First draw the diagram
Find the total resistance in parallel and later add the total resistance in parallel
1/R = 1/3 + 1/5 + 1/6
= 10 + 6 + 5
30
= 21/30
R = 30/21
= 1.43 ohms
total resistance in series = 7+8
= 15
total resistance in the circuit is 15 +1.43
= 16.43 ohms
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ELECTRICAL POWER IN CIRCUITS
Power = Energy used up
time taken
Electrical power = work done ( W) = V.I and is measured in watts
But we know that R = V/I and rearranging the formula it gives V= RI
substituting in the equation for V we get W = RI
2
and also I = V/R the substituting for I in the equation we get W = V
2
/R
Note
W = P.t when given time and power
W = VI is used when given voltage and current
W= RI
2
is used when given resistance and current
W= V
2
/R is used when given voltage and resistance
Example:
Find the electrical power dissipated in a resistor if the voltage supplied is 4V and the size of current
flowing is 3A.
Solution:
P = VI = 4V x 3A = 12 W.
Example 2. Resistance of 240 ohm and 120 ohm are connect in series then in parallel across 120
volts . Find
a. The current in each resistance?
b. The power dissipated in each resistance
In series circuit
Total resistance = R
1
+ R
2
= 120 +240
= 360 Ω
by ohms law. I =V/R
= 120/360
= 1/3
For 120 ohm resistor power = 120 x1/3x1/3
= 120/9
= 13.3 watts
For 240 resistor power = 240x 1/3x1/3
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= 240/9
= 26.6 ohms
In parallel circuit
The total resistance
1/R = 1/120 + 1/240
= 3/240
R= 80
Current = 120/80
= 3/2 A
In parallel circuit large current flow in small resistance therefore the two resistance will share each
other in the ratio of 1:2
Hence current in 120 ohm is 1A while in 240 is ½ A
But power = Voltage x Current
Power in 120 = 120 x 1
= 120W
In 240 ohms = 240 x1/2
= 120W
Or
We know that voltage is the same in any breach in parallel circuit
Current in 120 resistors = 120/120
1A
Current in240 resistor = 120/240
= 0.5A
Power = voltage x current
Power I 120 resistor = 120 x 1
= 120W
Power in 240 resistor = 240 x 0.5
120W
POWER TRANSMISSION
They two possible ways of transmitting power over long distance
1. By high current low voltage
2. High voltage low current
Transmitting at high current low voltage
When current is transmitted in such manner they high loss of power due to heating effect of the cable
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which will lost to sounding air in cables. The consumer will receive very few amount of power.
Transmitting at high voltage low current
The consumer receive high power because there is little heating effect in the cables and the drop is
very minimal.
Therefore to avoid power loss in cable power is transmitted at high voltage low current on a long
distance.
ELECTRICAL ENERGY:
Power = Energy / Time.
Energy = Power x Time.
But Power = VI
Therefore Energy = V. I.t used when given voltage and current plus time
or Energy = RI
2
t used when given resistance current and time
Energy = V
2
/R used when given resistance voltage and time
Example:
Find the amount of energy used by a bulb for two hours if the voltage supplied is 100V and the size
of current flowing is 5A for 2 hour .
Solution:
Energy = Power x Time .
But power = voltage x current = 100V x 5A = 500W.
NB Time must be in seconds.
Therefore Energy = 100V x 5A x 2 hours.
= 100V x 5A x 2hours x 60 minutes x 60 seconds.
= 3 600 000 joules.
Example 2. A heater is placed in 300g of water in a plastic jug. Find the temperate change if the
current following in heater is 10A and has a resistance of 5 ohm for 5 minutes
solution
Energy = RI
2
t
= 5 x 10 x10 x 5x60
= 15000 joules
H = Sm∆T
∆T = H/ Sm
= 15000/ 4.2 x 300g
= 119.05°C
COST OF ELECTRICAL ENERGY:
Electricity Boards use the kilowatt – Hour ( KWh ) as unit for charging electrical energy not joules.
1kWh is the amount of energy supplied in one hour to an appliance whose power rating is 1 Kw.
1 kWh = 3 600 000J
Example:
What is the total cost of leaving 2 light bulbs rated 100W each . switched on for 5hours at K1 per
unit?
Solution:
Energy = (2 x 100W) x 5hours =200W x 5hours = 0.2Kw x 5h = 1kWh
Therefore Total Cost = 1 unit x K1 = K1.00.
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HOUSE HOLD ELECTRICITY
In all houses components are always in parallel .The rating are as follows:
Lighting: The maximum current is 5A while for security light the fuses ranges up to 10 A
Sockets and heating equipments e.g. geysers, cooker, hotplates their fuse ranges from 13 A to 30 A
The three pin Plug
Live wire
It is the wire that caries current. It is the wire that gives shock when you touch it.
Neutral wire
It complete the circuit by attaching it to the earth by connects the wire to the earth
Earth wire
It protect the user from the Shock when you have touch by hand. The wire is connect to the earth
and metal body of appliance.
When they a shock the earth metal carries current from the body of the appliance to the earth since it
has low resistance rather than passing through you.
A fuse
It is a thin piece of wire which overheat melts if the current is too high. It is placed in the live wire
inside the plug. When current is too high it melt and break the circuit.
NB: other electrical appliances made of plastics do not have earth wire because the casing as the
earth.
Fuse value
The value tells the maximum current the fuse can allow.
Example
The following electrical appliances are connect in parallel, heater with three ring rated of 1.5 kW ,
woven with a rating 3.0Kw the Ac supply is 250V (2009 Maneb Revisited)
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Calculate:
1. The current through A and its resistance
The appropriate fuse for the circuit
Solution
1. for ring A
V= 250 , Power = 1.5 Kw , I =? R = ?
P = VI
I = P/V
I = 1.5 x1000/250
I = 6A
The resistance in A
by Ohm law R = V/I
R = 250/6
= 41.7 ohms
2. To find the appropriate fuse find the total power by adding
Total power = (1.5 + 1.5 + 1.5 + 3)Kw
= 7.Kw
But I = W/V
I = 7.5 x1000/ 250
= 30A
The best fuse should more that 30A but it should not exceed 35A.
MAGNETISM:
It is the study of behaviour of magnets
Magnets:
These are materials which attract other objects that contain iron or steel. Some magnets occur
naturally, while others are artificially made.
Man-made Magnets:
1. Alloy Magnets :
These contain metals E.g. iron , nickel, copper , cobalt and aluminum.
2. Ceramic Magnets:
These are made from powders .e.g. iron o
Magnetic force:
The force exerted a magnet is called Magnetic force or Magnetism.
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Magnetic Field:
It is the space or region under the influence of magnetic forces.
The magnetic field consists of imaginary field lines which originate from the north pole and moves
towards the south pole in form of circles.
The field lines are closest at the poles as such magnetism is strongest at the poles or ends.
Note:
The law of magnetic poles states that :
Like Poles Repel; Unlike Poles Attract.
Magnetization ( Making a magnet ):
It is the process of making a substance (magnetic material) to behave as a magnet
Magnets can be made in various ways such as through:
1. Stroking:
An object can be magnetized by being hit slightly by a permanent magnet. This make electron to
point in the same direction which make the metal to attract another magnetic materials. Here there
are two forms of stroking:
( a). Single Stroking:
In single stroking, one magnet is used to magnetize an object E.g. steel. Hitting starts from one
end and ends on the other end. During each time , the magnet is moved over certain distance. The
pole attained is opposite to that of a magnet used.
( b ) Double Stroking:
In double stroking , two permanent magnets are used ,but , stroking starts at the centre and
each magnet is moved outwards.
NB: It two like pole are used as follow . The magnet has the same pole on the ends as is said to
have consequent poles
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2. Electrical Method:
Direct current is allowed to pass through a solenoid with an object to be magnetized inside for few
seconds.
Demagnetisation ( Making a magnet lose magnetism )
It is the process of making magnets to loss its magnetic force
A magnet can be demagnetised in various ways. These include:
1. Ac. Method
The alternating current always changes its direction. This make the electron spins to keep on
changing direction or alignment inside in the wire, hence the object loses it magnetic force due to the
changing direction of electron spins since for object to behave as magnetic electron should point in
the same direction. While in direct current electron point in the same direction.
2. Heating: Another method of destroying magnetism is to heat the magnet to redness and then
allow it to cool while it is lying in an East- West direction. Heating cause electron to in excited
state and tend to be pointing in different direction. Since cooling is done while the magnet is
pointing in west and east direction this makes the magnet to lose it poles hence the material fails
to attain its pole and became demagnetised. Heat treatment will spoil the steel. So this method is
not recommended.
3. Hammering this break the magnet into small piece which make magnet to lose its pole
vibration or rough treatment such as dropping by stopping current to flow
ELECTROMAGNETISM:
It is magnetism achieved in a conductor when current flows. The magnetic field has the following
features.
1. The magnetic filed lines are in circles
2. The fields is strongest close to the wire
3. Increasing the current increase the strength of the field
How to increase magnetic force in a conductor
The magnitude of magnetic force in electromagnets can be increased by
1. By increasing the size of current
2. By wounding the wire into a long coil called solenoid.
3. By inserting magnetic material into the solenoid.
How to detect the direction of field lines.
They are two ways how the direction of field line can be detected
1. Maxwell s Screw Rule
The law state that imagine a right- handed screw being turned so that it bores its way along the wire
in the same direction as current The direction of turning gives the direction of the field.
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2. By right -hand grip rule
It state that imagine you are gripping a wire with you right hand and thumb point to direction of
current then the rest of finger point to direction of the fields.
Field due to a Current in a Solenoid:
The coil becomes as it has a N pole at one end and a S pole at the other
How to detect the poles of the solenoid
It can be detected using the right hand grip rule by imagining that you are gripping the coil with your
hand so that your fingers point in the convention current direction of current you thumb will point to
the north pole of the coil
Note:
Rule for the Polarity of a Coil Carrying Current:
When viewing one end of the coil, it will be of N pole if the current is flowing in an anticlockwise
direction, and of S pole if the current is flowing in a clockwise direction.
USES OD ELECTROMAGNETS:
They are used in:
i. Electric bell
ii. Magnetic relay
iii. Loud speakers,
iv. Electric motors.
v. In the ear piece of telephone.
vi. To lift heavy scrap metals.
FORCE ON A CONDUCTOR IN A MAGNETIC FIELD:
When a current –carrying conductor is placed in a magnetic field, a force is produced.
If a stiff copper wire is put at right angles to the field provided by U-shaped magnet , and current
allowed to flow through the wire ; the wire moves upwards. This indicates that there is upward force
acting on it. See the diagram below.
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How the force is created
The wire carrying conductor create a magnetic force in already existing magnetic filed by the
permanent magnetic. The two magnetic forces cross each other causing repulsion force which causes
the wire to be pushed away from the magnetic field cause by permanent magnet.
How to know the direction of force, current and magnetic field.
The direction of current ,force and the magnetic field can be determine by the Fleming's left hand
rule which states that the thumb finger point to the direction of force or thrust , second finger point
to the direction of current and the first finger point to the direction of field if the fingers are hold at
90 degree angle
Note:
1. If the direction of either the current or field is reversed , the wire moves downwards.
Electric Motor
It is made by two permanent magnet and a loop within the the magnetic field which is attached to
the direct electric current through two brushes made of carbon.
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How it works
When current flow into the loop for example through AB direction the wire is pushed upward
according to Fleming right hand rule by the magnetic fields set by the permanent magnets. Since it
is a loop the current also flow through from C to D which cause magnetic force to push the wire
down wards.
On the loop they are two force that is acting on it , the upward force on wire carrying into the
magnetic field (AB) and the force acting downward on the wire carrying current outwards.
Due to inertia, the wire overshoot and are reconnected to brushes and are re-pushed by the magnetic
force in the same way. The repeatedly pushing of the wire create a continuous movement of loop in
the magnetic field.
How to Increase the movement in electric motor
It can be increased by:
1. Increasing the number of turn in the coil
2. Increasing current
3. Increasing the strength of the magnetic field
ELECTROMAGNETIC INDUCTION:
It is the effect of producing electromotive force ( emf ) in a wire when the wire is moved across a
magnetic field.
The emf produced is present only while the wire is moving, and cutting through magnetic lines.
Similarly, when the magnet is moved to and fro in a coil, emf is induced in the coil
How to increase induced EMF (voltage)
1. Moving the wire at a higher speed.
2. Using a stronger magnet.
3. Increasing the length of the wire in the magnetic field. E.g. by looping the wire through the field
several times.(making more turns )
Faraday s Law of Electromagnetic Induction:
It states that :
The emf induced in a conductor is directly proportional to the rate at which the conductor cuts
through the magnetic field lines.
DIRECTION OF INDUCED CURRENT:
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The following are ways how the direction of induced current can be predicted
1. LENZ S LAW
The law helps to predict the direction of the induced voltage.
It states that induced current always flows in a direction such that it opposes the change producing
it. During production of induced current the direction of movement of a magnet or a solenoid change
the current always changes direction to oppose the change producing it.
2. The Fleming's right rule
It state that ,if the thumb, second finger and the first finger is ,mutually perpendicular to each other,
then if the thumb represent the direction of motion of and the first finger represent the magnetic
field , the second will represent the direction of induced current
This law helps to predict the direction of the induced EMF in a generator or a dynamo
3. Magnetic compass
The magnetic needle tend to point to the direction of the magnetic field set by the conductor
Like
4. Iron fillings
The tend to point to the direction of the induced magnetic filed set by the conductor
APPLICATION OF ELECTROMAGNETIC INDUCTION:
Electromagnetic induction is applied in:
i. Generators to produce electricity E.g. in bicycle dynamos HEP stations and other electricity
generators..
ii. Transformers to step up or down voltage.
The bicycle dynamo (electric Dynamo )
It is a device that convert the mechanical energy of movement to electric energy.
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How it works
The rotating coil cut the magnetic field lines set by the permanent magnet. this induces current in the
coil.
When the loop is horizontal lied it produced highest amount of voltage. When it is vertical the
voltage is almost zero.
In a bicycler dynamo ,the permanent magnet rotate in the creating moving magnetic field. This
changing magnetic field is being cut and re-cut by the permanent coil creating current
How to increase induced Emf in a dynamo
Induced Emf can be increase by:
1. Increase the speed of rotation
2. Number of turn in the coil
3. Increasing the strength of magnetic field
The difference between the electric motor and a dynamo( generators)
1. The electric motor used current to induce the movement in the coil while generator used
movement to induce current.
2. The generator or a dynamo uses slip ring while the a electric motor uses commutators.
TRANSFORMERS:
It is a device that acts a voltage multiplier. It change small amount of voltage to high voltage and
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high voltage to very low amount of current.
A transformer consists of a primary coil and secondary coil wound on a magnetic core.
The primary coil acts as the electromagnet, and the core guides the magnetic field so that field lines
pass through the secondary coil. It is used to change or transform ac voltage from one value to the
other.
How a transformer works
The ac supply is connected to the primary coil and emf is induced in the secondary coil.
The changing ac in the primary coil produces the changing growing and falling magnetic field that
cuts and re-cuts he secondary coil repeatedly- thereby producing an alternating current (ac ) in the
secondary coil
As current flows forwards and backwards through the primary coil , it sets up an alternating
magnetic field in the core. This induces an alternating emf and current of the same frequency in the
secondary coil.
If D.C supply is used, no emf can be produced in the secondary coil induced because direct current
does not set changing magnetic field that can be cut and re-cut by the secondary coil
Types of transformers.
1. Step –up Transformer
It has fewer turns in the primary coil, but has more turns in the secondary coil. It it act a voltage
multiplier by incrementing the amount of induced voltage
How it work
The small number of turns in primary coil set the rise and following magnetic field which are cut by
secondary coil. Since they more turn in the secondary this increase the number of cuttings per
second hence increasing the voltage in the secondary coil
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2. Step –down Transformer.
It has more number of turn in primary coil that in the secondary coil. It reduces the amount of
current from high to low voltage.
How it works
Since they more turn in the primary coil which set the rising and falling magnetic. The small number
of turns in secondary coil make some of magnetic filed line not be cut or lost this reduces the
induced current.
ENERGY LOSSES IN TRANSFORMERS:
Energy loss in transformers occurs due to:
1. The resistance of windings:
The windings of copper wire do have some resistance and heat is produced by the current in them.
2. Eddy currents:
The core itself is a conductor in a changing magnetic field. As the moving field lines of force from
primary coil cut the iron core , small currents are induced in the coil. These currents flow in small
circles, like eddies ( whirlpools ) in water when stirred. These currents are called Eddy Currents.
3. Leakage of Field Lines
All the lines produced by the primary coil may not cut the secondary coil, especially if the core has
an air-gap or is badly designed.
4. Magnetization and Demagnetization of the Core.
Work has to be done to alter sizes and direction of domains and heat is released in the process.
Ways of reducing energy losses in transformer
1. Energy losses due to resistance of winding can be minimized by keeping the coil resistance
low by making the coils from thick copper wire
2. These eddy currents are reduced by using a laminated core made of sheets, insulated from
each other to have a high resistance vanish
3. Energy losses due to leakage of field lines can be reduced by proper design of the
transformer.
4. Energy losses due to magnetization and demagnetization can be reduced by making the core
from soft magnetic material such as iron which is easily magnetized and demagnetized. E.g.
silicon, iron or steel alloy.
Note:
Laminated core: The core is made from thin sheets of soft iron which are insulated from each other
.e/g/ using high resistance varnish to cut down the eddy current flows. A core made in this way is
said to be laminated (layered )
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Note:
Very large power transformers have to be oil- cooled to prevent overheating.
TRANSFORMER CALCULATIONS:
Assuming the transformer is 100% efficient energy is conserved. Then:
Power in Primary coil = Power in Secondary coil.
i.e. V
P
x I
P
= V
S
x I
S
. where; V
P
= primary voltage; V
S
= Secondary voltage;
I
P
= Primary current; I
S
= Secondary current.
If current and voltage are to be on one side each; then rearranging, we have:
V
P
/ V
S
= I
S
/ I
P
. or I
S
/ I
P
= V
P
/ V
S
.
If you have being given voltage and number of turn the following formula is applicable
Secondary voltage(V
S
) = Secondary Turns ( N
S
)
Primary Voltage(V
P
) Primary Turns ( N
P
)
(V
S
) / (V
P
) = ( N
S
) / ( N
P
).
When you have been given current , and number of turn the following formula must be used
N
p
x I
p
= N
s
x I
s
Examples1
Calculate the number of turns on the secondary of a step down transformer , which would enable a
12V bulb to be used with a 240V ac mains power, if there are 480 turns on the primary.
What current will flow in the secondary when the primary current is 0.5A.Assume there are no
energy losses.
Solutions:
V
P
=240V; V
S
=12V; N
P
=480; N
S
? Using:
V
S
/ V
P
= N
S
/ N
P
; Then:
N
S
= V
S
/ V
P
x N
P
.
= 12V x 480 turns / 240V
= 24.
Therefore the number of turns on the secondary = 24.
2. V
P
= 240V; V
S
=12V; I
P
= 0.5A; I
S
= ? Using:
V
P
/ V
S
= I
S
/ I
P
;
Then:
I
S
= I
P
x V
P
= / V
S
.
=0.5A x 240V /12V
=10A.
Therefore the current that will flow in the secondary is 10A.
Example 2: A transformer has 2000 turns on the primary coil. The voltage applied to the primary
coil is 240V ac. How many turns are on the secondary coil if the output voltage is 48V ac?
Solution:
V
P
= 240V; V
S
= 48V ; N
P
=2000 ; N
S
= ?.
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Using:
(V
S
) / (V
P
) = ( N
S
) / ( N
P
)
Then:
N
S
= V
S
./ V
P
x N
P
= 48V x 2000 turns / 240V
= 400.
Therefore the number of turns on the secondary coil is 400.
Transformer efficiency
Efficiency is the ratio of the useful output to the input.
Transformer efficiency = output energy x 100%
input energy
Example; A transformer is used on the 240V ac supply to deliver 9.0A at 80V to a heating coil. If
10% of the energy taken from the supply is dissipated in the transformer itself , what is the current in
the primary windings?
Solution:
Power delivered to the heating coil = V
S
x I
S
.
= 9.0A x 80V
= 720W.
But power delivered to the coil = 100% - 10% = 90% because 10% is wasted in the transformer.
90% of power = 720W.
But input power = 100%.
Therefore input power = 720W x 100%
90%
= 800W.
Primary current = power / voltage
= 800W/ 240V
= 3.3A
ELECTRONICS
It the study of how tiny electrical circuits and devices that are used to make these circuits
BAND THEORY
This theory tries to explain why other substance conducts electricity while other do not conduct
electricity.
How bands are formed
When atoms are in gaseous state they are isolated and they occupy defined energy levels. As the
atoms are cooled, the atoms are they are brought together. Electrons wave from different atoms over
lap and the each energy level broadens into bands of energy which is shared by all atom.
There two bands that are formed and are separated by the gap called the forbidden gap
The conduction band
It is the out most bands. Where electrons go to.
Forbidden gap
It separates the conduction band and the valence band
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Valence band
It the inside band where electrons come from
Arrangement of bands in conductors ( metals)
In conductors E.g metal electrons are free to move since they are not bonded to any atom.
The slightest difference in metal make electron to move
They have smallest forbidden gap between the valence electron and the conduction band. This
makes electron easy to jump from the valence electron to conduction band making them good
conductors.
Arrangement of bands in semiconductors
They have a medium forbidden gap between the valence electron and the conduction band.
For the electron to move from the valency band to conduction band they need extra energy to lift the
electrons to the conduction band.
At absolute zero temperature the valence band is totally filled while the conduction band is empty
and no conduction can occur.
As the temperature increases the electrons can gain some extra energy to jump from the valence
band to conduction band making current to flow.
Arrangement of band in non metals
They have large forbidden gap and the valence band is full and the conduction band is empty . This
makes insulator not conduct electricity at any temperatures
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SEMI CONDUCTORS
They are materials that partly conduct electricity. Example of such material include, carbon inform
of graphite, silicon,
Types of semiconductors
They are two types of semiconductors,
INTRINSIC SEMICONDUCTORS
They are semiconductors naturally conduct electricity
Extrinsic semiconductors
They are semiconductors that conduct electricity when and impurity is added.
The impurity can be either a proton or an electron
TYPE OF EXTRINSIC SEMICONDUCTORS
Negative type semiconductors ( n- type )
The energy carrier is the negative electrons. An electron is doped into the natural semiconductor.
The impurity has an extra electron from a pentavalent element E.g Phosphorus, nitrogen . For
example an atom of phosphorus is introduced in the lattice of a semiconductor. The introduced
electron can be easily thermal excited to the conduction band making electricity to flow.
The positive type (p-types)
The conduction is done due to positive charge carrier. The impurity has a deficiency number of
electron compared to the semiconductor . For example boron is introduced into the lattice of group
natural semiconductor. This introduces an extra hole into the valence band. This hole is used to
transfer current.
NB the process of adding an impurity to a semiconductor is called Doping
Semiconductor diodes
They are made by joining the p -type semiconductor and the n -type semiconductor.
When the two semiconductors has been joined , the electron from the n-type are trapped by the p
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side holes and the conduction band has no electrons in it and also the p hole has also been filled.
This thin layer created is called the depletion layer. Charge carrier can not flow across the depletion
layer.
The positive type semiconductor makes the anode of the diode while the n type make the cathode of
the diode
Forward bias
The p-type should be connected to positive side of the cell so that the effect of the depletion layer
and n- type should be connected to negative side of the cell. In other word the anode should be
connected to the positive side of the cell while the cathode should be connected to negative side.
This makes the depletion layer to be canceled out and current should flow rapidly. The depletion
layer is canceled because like charges repel each other and unlike charges attract each other. Since
the negative semiconductor is connected to the negative side of the cell and the positive
semiconductor to the positive terminal of the cell this make the electron in n- type to repel with
other electron from the cell as a result the electron are attracted toward the depletion layer cancelling
it out. The diode in forward biased it conduct electron by electron in the n-type chip and by means
of holes in the p- type chip.
Reverse bias
The p type is connected to the negative side of the cell while the n-type is connected to negative side
of the cell or in other words the anode is connected to negative side of the cell and the cathode is
connected to the positive side of the cell . This increases the depletion layer because the p types
contain the negative ions. These negative ions prevent the electron to enter the depletion layer and
no current can flow
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Types of diodes
Light limiting diodes LED
They are made by gallium arsenide phosphide. It emit red, yellow or green light depending on ton
its composition.
Its uses
1. They are used as numerical display in electronic calculators, clocks cash register and
measuring instruments.
2. They are used as indicators in electrical equipments .it show is current is following into
the equipment E.g. in computer and radios
3. The are use as bulb of LED torches ( china torches)
Zener diode
They are made up of silicon junction.
Uses of Zener diodes
1. They are used as power stabilizer in circuits by making current to flow in one
direction
2. They are used as rectifiers where it changes Ac to direct current
3. They are used in Analogue Modulator radio when combine by a capacitor to act as a
tuner.
Photo diode
It has a transparent window through which light can enter
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Uses
They are used as optical fiber receiver which generates a pulse of current every time a beam of light
is interrupted.
TRANSISTORS
They are made up of two type p-n junction in the same crystal. They are made up of germanium
because it with stand high temperature but due to easy mass production of silicon n n-p-n diode the
germanium is less preferred. It uses three connections, the base, the collector and the emitter. When
connected to a circuit, current enters through the emitter and leave through base but convectional
current enter through the base and collector and leave via the emitter.
Types of transistors
Transistors are grouped depending on the arrangement of the p-n junctions
P-N-P JUNCTION TRANSISTORS
They made in such way that the middle chip that form the bases is the semiconductor of the n type
while the emitter and collector are made up of the p- type
The n-p-n junction diode
The middle plate is the p-type while the two end plates are the n-type
NB. The arrow in transistor symbol shows the direction of convection current.
USES OF TRANSISTORS
They are used as voltage amplifier
The transistors combine by three resistors or a small current is introduced through the base as shown
below
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When a small current is introduced at the base the big signal comes out from the collector circuit. In
both circuits Ib + Ic = Ie which is current which has be amplified. The resistors prevent large current
to pass through the base rather it should pass through the collector via the emitter and only small
current should pass through the base.
They are used as a switch
When a transistor is combined with a light sensitive resistor or thermostats it act as a switch the
circuit is set like follows
When the light intensity increase the resistance increases stopping current to flow and switches off
the collector remitter current. When the light intensity decrease the resistance switching on the
collector remitter current.
When the temperature increases the resistance increases stopping current to flow and switches off
the collector remitter current. When the temperature decrease the resistance switching on the
collector remitter current.
Transistor as potential divider
The circuit is set as follows
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The emitter base Voltage can be changed from zero to maximum values. When emitter base voltage
is at zero the no current flow through the transistors. But when the emitter base voltage is at
maximum the current flow the transistors. In this way the transistor acts as a potential divider.
THERMISTORS
It is a semiconductor device that its resistance changes as the temperature changes.
The resistance decreases as the temperature increases.
They are use as temperature change sensor in electronic equipment
They also use as electronic thermometer in refrigerators
LIGHT DEPENDENT RESISTORS
Their resistance decreases as the intensity of light decreases.
Uses
They are used in photographic expose meter
They are use in series with the resistors to provide an input for transistor or in logic gates
CAPACITORS:
A capacitor is an electronic device that stores charge.
It consists of two parallel plates separated by an insulator called dielectric.
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Note: There are several types of capacitors.
CAPACITANCE ( C )
CAPACITANCE ( C )
It is the ability to store charge.
Capacitance is given by the formula: C = Q/ V:
where C = capacitance; V = voltage in volts; Q= charge;
Capacitance increases if :
i. The area of the plates is increased.
ii. The separation between the plates is decreased.
iii. If the insulator (dielectric) is a solid and not air.
USES OF CAPACITORS:
1. They are used to separate DC signals from ac signals E.g. as used in radios, TVs and computers.
2. They are used to smoothen (filter) DC signals in rectifiers used in radios, TV and computers.
3. They are used to filter frequencies E.g. in radio and amplifiers in tone control circuits.
4. They are used in some cameras for flashing when taking photographs.
Exercise
2003
Question 1
a. Figure 1 is a diagram of electric circuit
i. What kind of energy is produced in 12Ω resistor as electric current flows through it?
ii. Calculate the amount of energy produced in the resistor if the electric current flow
through it for 10 minutes
Question 7
a. Briefly explain how a piece of iron is magnitised by stroking
b. (i) describe how a step up transformer works
ii. Explain two ways in which energy in transformer loss are minimized
2004
a. Explain why voltmeter is connected in parallel while an ammeter is connected in series
b. Figure 2 is a diagram of resistor
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Brown = 1 Green = 5 Black = 0
Gold =5%
What is the value of the resistor?
c. (i) Explain the difference between an electric motor and an electric generator
ii. State any two factors which affect the amount of voltage produced by a generator
d. (i) What is a semiconductor?
ii. Draw a circuit diagram in which a bulb , a cell and a diode are connected in series such that the
diode is forward biased
e. What is meant by doping
f. State two function of transistors
g. (i) is the head lamp of a motor car connected in series or parallel
ii. Give a reason for your answer to 4gi.
2005
Question 1
d. Figure 2 is a diagram of a compass needle placed under a connecting wire
i. Draw the arrow on the diagram to show current and compass direction when stwitch is
closed
e. Mention any two devices that use electromagnets
Question8
a. A 6kw power is feed to transmission cable of a resistance 3Ω. Calculate the power wasted in
cables if the power transmit 300V
b. Describe an experiment that could done to find out if electrical resistance of the wire varies
directly with length
3PPII
a. With an aid of a labeled diagram explain how a step down transformer works. in your
diagram , show the coil in primary and secondary coil and current source
b. Explain how fuse works
2006.
Question 4
a. Figure 3 is a circuit diagram
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i. Name the components labeled X and Y
ii. Calculate the total resistance
iii. Calculate the current in the 30Ω resistor
iv. Work out the voltage across the 10Ω resistor
b. (i). Define kilowatt – hour
ii. The power rating of a television is 150W. How much power in kilowatt- hour will use if it
runs for 10hour
iii. If the cost for power is K5.00 per kilowatt hour, what will be the cost of running the
television in 4dii. For 10 hours per day for 2 days
Question 3 ppII
Describe an experiment that could be carried in order to determine the length of nichrome wire that
could be used to make 1.5 resistors
2007
Question 4
b. A current of 2A flows through an electric heater connected to a voltage of 240 V. calculate the:
i. Resistance of the element
ii. Power dissipated by the heater
iii. Cost of running the heater for three hours if the cost of electricity s K5.00 per Kw-hr
c.(i) Give two advantages of alternating current over direct current.
ii. State any two causes of energy loss in transformer
iii. How can each energy loss mentioned above can be minimized
d. Figure 2 is a diagram of electric circuit
i. If S1 is closed and S2 open, calculate the reading of the ammeter and voltmeter
ii. What will be effect on the voltmeter reading if both switches are closed?
iii. Give a reason for your answer
Question 4 PPII
With an aid of diagram, describe an experiment that could be done to identify unknown substances
W, X and given that are diode, and insulator, a resistors but not necessary in that order.
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2008.
Question 4
a. Figure 2 is a diagram of transformer
i. Name the type of transformer shown in the diagram
ii. Give a reason for your answer to 4ai
iii. Calculate the out put of the transformer
b. (i) what is a semiconductor
ii. Explain how the temperature of semiconductors affects their electrical conductivity
c. Figure 3 is a diagram of a transistor
i. Name the parts marked X and Y
ii. Give any two uses of transistors
d. i. Why diode some times referred as rectifier?
ii. Draw a circuit consisting of cell bulb and a diode such that it is forward biased
e. Figure 4 is a diagram showing a charged rod brought to a metal can standing on an insulator
i. complete the diagram by indicating the induced charges on the can
ii. Why was the can placed on an insulator?
2009.
Question 4
a. What do the following electrical symbols stand for
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b. In a circuit a 6 Ω resistor and 3 Ω resistor are connected in parallel and 8Ω resistor is
connected in series with them.
i. Draw a circuit diagram using the given information
ii. Calculate the total resistance in the circuit.
c. Figure 5 is a diagram of electric circuit for a house
i. What type of circuit is shown in the diagram
ii. Give a reason why this type of circuit is preferred for wiring house
iii. Explain the importance of including device X in the circuit in the figure
iv. Give one appliance where device X is used
d. Figure 6 is a diagram showing colour codes and the resistors marked P and Q
i. What is the resistance of resistor P
ii. What is the advantage of using resistor P in a circuit that resistor Q
2010
Question 6.
a. Mention any two appliances that uses the principle of electrostatics
b. Figure 4 is a diagram of electric circuit.
Colour
Number
Black
0
Brown
1
Red
2
Orange
3
Yellow
4
Green
5
Tolerance
Gold = 5% silver = 10%
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i. Name the parts labelled X
ii. What would happen if the part labelled X is brought closer to iron fillings
iii. Give a reason for your answer to 6bi.
iv. What is the role of iron bar in the circuit
c. Define electrical power.
d. An electric bulb is marked 240V , 60w
i. What is the meaning of 240, 60W
ii. Calculate the heat dissipated by the bulb, in joule , if it operate for 8 minutes
e. A step down transformer has 1200 turns in the primary coil and 50 turns in the secondary
coil. Calculate the voltage in the secondary coil if the voltage in the primary coil is 240V
f. Explain how the transformer works
2011
Question 2
a. Give one type of semiconductors
b. Explain why the difference in total current between similar resistor connected in parallel and
in series
c. Give one method of demagnetising permanent magnet
d. Figure 2 is a diagram showing an experiment on electromagnetism in which a current
carrying wire passes through the centre of the card board on which iron filling were sprinkled
i. Draw an arrow in the compass to show the direction of field lines
ii. What would happen to the direction of compass needle if the direction of current is
reversed
iii. Give a reason for your answer
2012
Question 2
a. State two ways of inducing electromotive force ( Emf)
b. Mention any one application of electromagnetism
c. Figure 2 is a diagram of an electric circuit
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i. What is the function of the part labelled Z
ii. State any two advantages of the circuit in the figure above
iii. Which bulb would use more current if two switches were closed
iv. Give a reason for your answer
d. Calculate the power dissipated in an electric heater in which 4A of current flows when
connected to a 230V supply
8e. explain how a piece of a steel could be magnetised by single touch stroking method .
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NUCLEAR PHYSICS
ISOTOPES
They are atom of the same element that have the same number of proton but different number of
neutrons or are elements with the same mass number but different atomic mass number
For example hydrogen exists in three forms namely. Protium, deuterium and tritium
Name
Number of neutrons
Symbol
Representation
protium
0
1
1
H
Hydrogen -1
deuterium
1
2
1
H
Hydrogen-2
Tritium
2
3
1
H
Hydrogen- 3
NUCLIDE
Are atoms with which have different nuclei. This different may due to different in atomic mass,
atomic number or number of protons.
All isotopes are nuclide
TYPES OF NUCLIDE
Nuclides are categorized according to their stability. They are two type of nuclide.
STABLES NUCLIDE
Are nuclide s that does not undergo spontaneous decay to became the nucleus of different elements
UNSTABLE NUCLIDE
Are nucleus that spontaneously undergoes decay to become the nucleus of different elements
Unstable nucleus undergoes the process of transmutation and radioactive decay.
TRANSMUTATION
It is the process where by the nucleus of an atom undergoes transformation into a nucleus of
different elements
In transmutation a nucleus can gain or loose a proton to become a nucleus of another element
For example carbon 14 nucleus can lose an electron and change a neutron into a proton and become
Nitrogen 14
RADIOACTIVITY
It is the ability of unstable nucleus to undergo spontaneous decay.
NB: The radioactive active decay is said to be spontaneous because it is not affected by
environmental factors such as temperature, pressure, or presence of catalyst. It is not dependent to
other chemical reagents or physical factors being present. It takes place without any intervention.
WHY ELEMENT UNDERGO RADIOACTIVE DECAY
To gain stability since its elements ant to be stable
To release extra energy from the nucleus
CHARACTERISTICS OF RADIOACTIVE SUBSTANCES
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1. They continuously undergo decaying into simpler atoms as a results radiation
2. Radiations from radioactive substance produce bright flashes of light when it hut a certain
compound.
3. They cause ionization of air molecules like they knock out electrons from molecules of air
leaving with positive charge
4. Radiations from radioactive substance can penetrate the photographic film leaving it black
5. Radiations from radioactive substance can destroy the germination power of seed of plants , kill
bacteria and burn or kill animal cells and plants.
TYPES OF RADIOACTIVE DECAY
E
1. Natural radioactive decay
It is radioactive decay that occurs naturally without the aid of human. Tit occurs in naturally
existingelements radioactive elements such as -uranium -polonium –radon –thorium.
NB All elements beyond bismuth in the Periodic table are radioactive.
ARTIFICIAL RADIOACTIVY
They occur by human aid where atomsare slowly bombardment in the nuclear reactor.
A proton or a neutron is used to bombard the nucleus of a stable nucleus. This makes the nucleus to
become unstable and undergoes nuclear decay
RADIATION
It is a particle or energy released due to decay of a radioactive substance.
TYPES OF RADIATIONS
These include: alpha particles, beta particles and gamma rays .
ALPHA PARTICLES (α - Particles )
An alpha particle is a nucleus of Helium atom and is made up of two protons and two neutrons.
When an atom decays by alpha particles - emission its nucleon number decreases by 4 and its proton
number by 2.
For example
Radium decaying to radon.
226
88
Ra
222
86
Rn+
2
4
H
Properties of alpha particles:
1. They carry a positively charge because they contain protons without electrons
2. They all travel at approximately the same speed 1/10 the speed of light.
3. Thy are the most strongly ionizing power of the nuclear radiations
4. They tend to attract electrons away from nearby atoms.
5. They are the least penetrating- they can be stopped by a thick sheet of paper.
6. Collisions with atoms finally brings them to rest
7. They can be deflected by magnetic and electrostatic fields.
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BETA PARTICLES (β-particles )
In a beta –decay,a neutron changes to a proton and an electron. The proton remains in the nucleus
and the electron is emitted as a beta- particle.
The proton number increases by 1 since it has one more proton.
But, the new nucleus has the same mass ( nucleon ) number. E.g.
Radioactive carbon called carbon-14, decays by b- emission to nitrogen .
14
6
C
14
7
C +
-1
e
0
In beta decay a proton in the nucleus is converted to a neutron and a positron .E.g.
64
Cu
29
64
Cu
28
+
+1
e
0
NB. Positrons are subatomic particles with the same mass as an electron , but positive charge . The
symbol for a positron is
+1
e0
NB. Beta decay can either release a positive particle or a negative particle. If the decay release a
positive particle a positron the decay is know as positive beta decay while if and electron is released
the decay is callednegative beta decay.
Properties of beta particles:
1 They are electrons emitted with varying speeds which can be high as 9/10 of the speed of light .
2 They have e negative charge .
3 They are lighter than alpha particles .
4 They can be stopped by a few mm of aluminum or Perspex
5 They are less ionizing power compared to alpha particle
6 They are more penetrating.
7 They show much greater deflections than alpha particles because of their lower mass.
( g- ) GAMMA RAYS
These are electromagnetic waves of very short – wave length.
They result from energy changes inside atomic nuclei, as distinct from X- rays which are caused
by energy changes outside .
A gamma photon is emitted at the same time as an alpha or beta particle.
After emitting alpha and beta particles , some nuclei are left in an excited state. Rearrangement of
the protons and neutrons occurs and a burst of gamma rays is released.
Properties of Gamma rays
1. The are the most penetrating .Their intensity is greatly reduced by several cm of lead , but are
never completely absorbed .
2. They are the least ionizing
3. They are not deflected being electromagnetic waves or electric field
COMPARISON OF ALPHA, BETA AND GAMMA RADIATIONS
RADIATION
S TYPE
IONISING POWER
PENETRATIN
G POWER
RANGE
IN AIR
(m)
ELECTRIC
CHARGE
ABSORBED BY
ALPHA
high
Low
0.05
+2
Paper
BETA
medium
Medium
3
-1
5mm aluminum
GAMMA
low
High
100
0
3cm lead halves
intensity
DEFLECTION OF NUCLEAR RADIATIONS IN MAGNETIC FIELD
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Like any stream of moving charged particles,alpha and beat particles are deflected in magnetic or
electric fields .Beta particles show much greater deflections. Gamma rays are not deflected being
electromagnetic.
Penetration power of radiation on human body
NUCLEAR EQUATIONS
Are equation that represent the decay process of a nucleus.
In nuclear equations, the following point should always be remembered
1. In any nuclear equation, the total mass number on the left hard side should be equal to the
atomic mass in the left hand side
2. In any nuclear equation, the total atomic number on the left hard side should be equal to the
atomic number in the left hand side
3. Determine the decay products that will be produced a balanced equation
4. Write the balance equation
5. Don't concentrate on the charge because in nuclear physic we are concerned with nucleus not
the atom
EXAMPLE :
What is products are formed when a neutron is bombards a nitrogen atom?
14
7
N +
1
0
N
14
6
C +
Y
X
total atomic mass number 15 = 14+ 1 so x = 1
total atomic number 7= 6+y so y =1
So the equation is
14
7
N +
1
0
N
14
6
C +
1
1
H
DETECTING NUCLEAR RADIATIONS:
Most methods of detecting alpha, beta particles and gamma rays are based on the fact that that
these have an ionising effect .These include :
PHOTOGRAPHIC PLATES
The photographic film is blackening when exposed to amount of radiation. The amount of
blackening determines the level of exposure. This method is used by people who work with radiation
to check the amount they are exposed to.
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IONIZATION DETECTOR
The ionization detector detect the amount of air ionized by the radiations and measure the amount of
radiation
SPARK COUNTER
It has positively charged wire grid and the negatively charged metal plate. When the radiations pass
between them it ionizes and the electron which cause large current to suddenly pass through the air
under high voltage causing a spark which is heard or registered by the electronic device
CLOUD CHAMBER
It enables tracks of charged particles to be seen .
When air containing vapour, E.g. alcohol , is cooled enough , saturation occurs . If ionising
radiation passes through the air, further cooling causes the saturated vapour to condense on the air
ions created . The resulting white line of tiny liquid drops shows up as a track when illuminated .As
the alcohol vapour spreads downwards, it is cooled
Each time an alpha particle is shot ,it ionises the air in its path and alcohol readily condenses around
these ions .A narrow cloud made up of millions of tiny alcohol droplets forms along the track of
alpha particle..The track is visible because it reflects light from the lamp
BUBBLE CHAMBER
Radiation leaves a trail of bubbles in liquid hydrogen .
A magnetic field is usually applied across the bubble chamber which causes charged particles to
move in circular paths .
The higher density of atoms in the liquid gives better defined tracks, than obtained in a cloud
chamber .
ELECTROSCOPE
A charged electroscope discharges a radiation source is brought near the cap.
Radiation from radium knocks electrons out of surrounding air molecules leaving them as positively
charged air ions .The air ions are attracted to the cap or repelled depending on the charge on the the
cap. As a result in either case, the charge on the electroscope is neutralize or lost
GEIGER –MULLER TUBE
When radiation enters a GM tube , either through a thin end window made of mica or through the
wall if it is very penetrating , it creates argon gas ions and electrons . These are accelerated towards
the electrodes and cause more ionisation by colliding with other argon gas atoms .
On reaching the electrodes , the ions produce a current pulse which is amplified and fed to either to
a scale or rate meter.
The scale counts the pulses and shows total received in a certain time.
The rate meter gives the counts per second or minute, or count rate directly. It usually has a loud-
speaker which gives a click for each pulse.
SCINTILLATION COUNTER
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It consists of fluorescence crystal placed in contact with a photomultiplier. When ration strike the
crystals , light is emitted by the photomultiplier and amplified by electronic circuits and is shown
on the fluorescence screen
NUCLEAR REACTIONS
Nuclear reaction is any process in which a particle penetrates a nucleus and changes it in some
way .
Nuclear reactions involve rearrangement of protons and neutrons in nuclei so that new elements are
formed.
In a nuclear reaction the total mass of all the product particles is slightly less than the total mass of
the reactant particles. This loss of mass appears as a large quantity of energy released in form of
heat.
Uses of nuclear reactions:
1. For the provision of energy to make electricity.
2. For destructive purposes – Nuclear bombs release large amounts of energy .
3. Used to generate power heart pacer machines
Types of Nuclear Reactions
A NuclearFission
It is the splitting of an atom of a heavy element into atoms of lighter elements , with the release of
energy
Fissile isotope is bombarded by neutrons. As a result of interaction between neutrons and atoms of
the isotope , fission of some atoms occurs.
More neutrons are produced by the nuclear fission than are absorbed .
If neutrons from the fission of one nucleus go on to split other nuclei , a chain reaction may result .
In practice , neutrons are lost by escaping without causing further fission .
This must exceed a certain critical value to sustain the chain reaction. This is called critical mass .
In other words critical mass is the minimum mass of a radioactive element E.g. Uranium –235
necessary for any atomic explosion.
Example of Fission reaction
Uranium –235 can split into barium and krypton with 2 or 3 neutrons.
Uranium
atom
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92
U
235
+
0
n
1 90
Kr
36
+
56
Ba
144
+ 3
1
n
0
.
NB . The barium and krypton nuclei are themselves radioactive .
Nuclear Reactors
They consist of a pressure cylinder containing the following :
Nuclear fuel elements :
The fuel is made from uranium dioxide .
Graphite core :
It acts as a Moderator
This slows down fission neutrons for a chain reaction to be maintained . This is because slow
neutrons are more effective in causing fission , unlike fast moving neutrons which are not absorbed
by uranium 238 .
Control rods :
These are boron –steel control rods . The boron absorbs neutrons .
Fission rate is controlled by raising or lowering boron –steel control rods .
If the rods are raised , more neutrons are able to cause fission , and the temperature rises .
The reactor can be shut down by keeping the rods in the lowered position .
NB Carbon dioxide gas is pumped through the core and carries off heat to the heat exchanger where
steam is produced .
Fusion Reaction
It is the combining of lighter nuclei to form heavier elements .
Heat energy is supplied first . Therefore the reaction is also called Thermonuclear reaction.
A lot of heat energy is generate compared to the heat released in fission reaction .
Example of fusion reaction :
Isotopes of hydrogen ( Deuterium and Tritium ) combine to form Helium .
2
H
1
+
3
H
1
4
H
2
+
0
n
1
NB . The sun is powered by fusion reaction . High energy converts hydrogen into helium .
Research is being carried on fusion reaction to find out if it can be used to generate power
RADIOACTIVE DECAY : HALF –LIFE ( DECAY RATE )
Half –life of a radioactive substance is the average time for half the atoms in a given sample to
decay.
The rate of decay is unaffected by temperature or chemical change .
Half life equation formula
The rate of decay is fixed for any given nuclide and is directly proportional to the number of nuclei
present.
Given mass of the a substance use the proportionality to find the half life of the substances.
Example
A radio active substance has the half life of 20 minutes. What fraction of the substance will reaming
after 1 hour?
After 20 minutes fraction left = ½
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After 40 minutes fraction left= ½ x ½ = ¼
after 1 hour fraction left = ½ x ¼ = 1/8
Example 2
Quarter of the uranium sample present in rock. Find the age of the rock assuming that uranium has
half life of 4500million years
Solution
After 4500million the fraction remain is ½ of the sample
There fore ¼ of the sample it means we should multiple the half life by two to get the age of the
rock
2 x4500 = 900 million years
ACTIVITY:
It is the average number of disintegrations (i.e. decaying atoms ) per second .
Half –life is also the time taken for the activity of a given sample to fall to half its original value .
Half –life or Activity can be measured by finding the count rate using GM tube .
Half –life vary from millionth of a second to millions of years .
Element or nucleus Half-life
Tin 1 700 000 000 billion years
Thorium 14 billion years
Uranium 4.5 billion years
Carbon 5566 years
Radium 1622 years
Scandium 85 days
Barium 86 minutes
Nitrogen 7.4 seconds
Helium 2.4 x 10
-21
seconds
DECAY CURVE
It is a curve that illustrates how elements decay .This can be used to show the rate of disintegrations
or decay e.g.
APPLICATIONS OF RADIOACTIVITY
It has a large number of uses in industries .These include :
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Medical uses
1. Treating cancer by subjecting a patient to controlled amounts of gamma rays from a Cobalt –60
source.
2. Used in studying the function of thyroid gland by drinking iodine-31
3. To detect blood clot in the body where the patent is injected sodium 25 into the blood system and
detect where it stops, tracing brain tumors and also to detect internal bleeding.
4. It help to make picture of organs in the body by use of gamma camera and are later detected
gamma ray from the pictures what form a picture on the screen
5. Sterilising equipment and instruments using gamma rays.
Industrial use
1. Controlling thickness of paper , rubber , metals and plastics accurately
2. Controlling the filling of packets and containers.
3. Tracing fluid flow in pipes .
4. Used in textile industrial where beta radiation fix various chemical onto cotton fibre
5. Used to trace the flow of liquid in chemical factory
6. Used to preserve food where the food is exposed to gamma ray
7. Used to detect hidden flaws in metals castings
Agriculture use
1. Tracing uptake of fertilizers by plants .
2. They are used to control the screw worm fly pest in South America
3. It help to reveal the rate of photosynthesis by growing plant in the air containing carbon 14
NB isotope with short halve life is used in agricultural studies so that the before the plant
became mature and ready to by consumed the radioactive decay has already completed and it can
not harm the consumer
Energy Source
1. Used in generation of electricity. Energy produced by radioactive fission of uranium-235 is
used in nuclear power station to provide electricity.
2. Used as source energy in artificial peacemaker for heart
3. generate power
Radio dating ( Archeology )
Carbon 14 is used in dating in Archeology
NB The uses of a radioactive isotope depend on its half –life . If the half –life is too short or too
long it can have little practical application.
Dangers and Safety:
Gamma rays are the most dangerous because they can penetrate to tissues deep in the body
The danger from alpha particles is small , unless the source enters the body .
1. The ionising effect produced by radiation causes damage to cells and tissues and can also lead to
the mutation of genes and deformity .
2. Beta and gamma radiations can cause radiation burns i.e. redness and sores on the skin and
delayed effects such as : Eye cataracts - cancer
3. Large exposure can lead to death .
4. Damage to blood system may lower the resistance to some diseases.
5. It can sterility
Safety
The following safety precautions:
1. The source should be handed with forceps provided and never touch with hands
2. They should never be pointed towards a person
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3. Food should not be taken where the source are being used because it may contaminate food.
4. Never smoke near radioactive source
5. wear rubber gloves , and hands should be washed after the source has been put away safely
6. Personnel operating nuclear reactors and the surrounding communities are protected from
excessive radiations by proper shielding –steel and other operational safe –guards .
7. Wastes of nuclear reactors are stored in containers in caves and abandoned mines . Some
containers are sunk in the sea .
Exercise
2003
Question two
a. What is an isotope?
b. Chlorine has two isotope
35
17
Cl and
37
17
Cl.
i. Give the number of neutron in
35
17
Cl
ii. Given that two isotopes are present in ordinary chlorine in the ratio of three to
35
17
C l
one atom of
37
17
Cl. . Calculate the atomic mass of chlorine
c. How would the chemical properties of the two isotopes compared? Give a reason for your
answer.
d. (i) What is radioactivity
(ii) When the nucleus of radium emit an alpha particle it decays to radon according to the
following equation
226
88
Ra
222
86
Rn +
4
2
He
1. Beside radon, name the other particle is produced when radium decay?
2. How does the mass of decaying atom compared with the masses of the products?
3. Name the alpha particle in this equation
e. Explain how the gamma rays are emitted
f. Give one sources of gamma rays
2004
Question 6
a. State any two safety precautions when handling radioactive substances
b. (i) describe “alpha particle” and “beta particle”
(ii) Explain why gamma radiation is used sterilization of medical equipments
c. (i). the radon-222 isotope is formed from beta decay of uranium -226(Ra). Write nuclear
equation of the reaction
ii. Figure 4 is a diagram showing a radiation passing through an electric field
Draw and label in the diagram the apath taken by alpha and beta – particles
iii. Explain the behavior of
d. The particles shown in the diagram 6 c ii
2005
Question 4
c(i). The half life of a radioactive substance is 3 hours. What mass of the substance would remain
after 12 hours if the initial mass was 20 g
ii. Why it is important to muse radioisotopes with shorter half life as tracers in agriculture.
2006
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Question 3
a. Define nuclear fusion
b. Name the two particles that are found in the nucleus
c. Two radioactive samples showed the following characteristics:
Effects on gold leaf
electroscope
Effects on card board
A
No effect
Sample passed through
B
Discharged
Sample blocked
i. Identify samples A and B
ii. Why does the particles emitted by B get blocked by card board
d. Figure 1 is a diagram showing radiation passing through an electric field
i. Name the particle taking path X Y and Z
ii. Explain why particles Z will be deflected towards the positive plate
2007
Question 8
Describe the following radioactivity process: alpha decay, beta decay and gamma emission . in your
description include atomic number penetrating power , ionising ability and behavior in magnetic
and electric field
2008
Question three
a. Define nuclear fusion
b. The equation below shows the fusion of heavy hydrogen
2
1
H +
2
1
H
3
2
H +
1
0
n
i. Name the products
ii. Identify the type of fusion
c. Uranium
239
92
U decays in the series of stages as follows
Stage 1
238
92
U
234
90
Th
Stage 2
234
90
Th234
91
P
i. Which particle is emitted at each stage?
ii. Apart from the particles mentioned in 3c I what else is emitted at each stage
d. State three safety precaution that must be followed when handling radioactive substances
e. When a source of radiation id placed in front of Geiger Muller counter, the initial count is 128.
After 16 minutes, the count rates is 8. Calculate half life of the source.
2009
Question three
a. Define radioactivity
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b. Name the three types of radiation
c. Mention any instrument that are used to detect radiation
d. (i)How does the chemical properties of the isotopes uranium
238
92
U and
234
92
U compared
ii. Give a reason for you answer in 3di
e. Mention any one natural source of radiation
f. A radioactive source has a half live of 30 minutes. Calculate the fraction left after 2 hours
2010
Question 5
(ai). Name two particles found in the nucleus of an atom
ii. State two characteristics of radioactive substances
b. State any two uses of nuclear radiation
c.Table 1 shows the activity of radioactive element with time
Activity (disintegration per second)
40
30
20
10
Time ( days)
0
4
8
16
i. Plot the graph of activity against time
ii. Use the graph to find the time taken for the activity to drop from 20 to 15 disintegration
per seconds
iii. Name the particles emitted when radium
236
88
Ra decays to (
222
88
Ra)ear f
2011
di. Define nuclear fission
ii. Write two symbols representing an isotope of hydrogen
e. Explain why charged electroscope is not suitable for detecting beta decay
f. Thorium (
232
90s
Th ) decays by alpha emission to radium (
228
88
Ra)
1. What do numbers 228 and 88 represent in
2. Write down the equation for the decay of Thorium
2012
6d.
i. Name any two types of radioactivity
ii. Define half life of a radioactive element
iii. Mention any three properties of an alpha particle
8.d. Eexplain any two uses of nuclear radiation
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REFERENCES
1. Abbott A.F (1989) 5
th
ed. Physics Heinemann Educational publishers.
2. Chicago Bruce Banyard and etl (1983) Fundamental Physic. Long-man Publication. Cheshire.
3. Child Ann (2000) Macmillan Secondary Chemistry . Macmillan Publisher Limited Malaysia
4. Earl B. and Wilford LDR (2004) 2
nd
ed . GSCE Chemistry John Murray publisher Limited
London
5. Duncan Tom, 2
nd
ed. GSCE physics for Today a d Tomorrow. John Murray Publishers.
6. Freemantle Micheal (2000) Chemistry in Action Prosperous Printing Co. Ltd HongKong.
7. Gallagher Rose and Ingram Paul . (2000) Complete chemistry . Oxford university press .New
York
8. Gerrish h. Howard, (1968) Electricity And Electronic. The Good Heart – Wilcox .Co. Inc
Illinois
9. Hill Graham and Holman John (2000) 5
th
ed. Chemistry in Context , Nelson Thones ltd
Cheltenham
10. Malawi institute of education Malawi school teaching syllabus Zomba
11. Maneb Unpublished past paper
12. Pople Stephen (2001) Complete physic , Oxford University press Milan
13. Rowel Gilbert and Sydney Herbert (2004) Physic. Cambridge university press. Cape town
14. Salvatore Tocci and Claudia Viehlland ,(1998) Holt chemistry visualizing matter. Holt
Rinehart and Winston publication Chicago
15. Taylor Micheal (2008) Macmillan Secondary Physics, Macmillan publishers Ltd Oxford
16. Wallis Keith . 3
rd
Chanco Physical Science for Malawi BK2 Chancellor College Publication
Zomba
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nd
Physical Science for Malawi BK1 Lilongwe
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